1. Energy & Chemistry 2H 2(g) + O 2(g) → 2H 2 O (g) + heat and light This can be set up to provide ELECTRIC ENERGY in a fuel cell. Oxidation: 2 H 2 → 4 H + + 4 e - Reduction: 4 e - + O 2 + 2 H 2 O → 4 OH - H 2 /O 2 Fuel Cell Energy

2. ENERGY is the capacity to do work (w) or transfer heat (q). HEAT HEAT is the thermal energy that can be transferred from an object at one temperature to an object at another temperature – Net transfer of thermal energy stops when the two objects reach the same temperature. Other forms of energy — 1.Radiantlight) 1.Radiant (light) — energy in light, microwaves, and radio waves 2.Thermal kinetic and potential 2.Thermal (kinetic and potential) — results from atomic and molecular motion – Temperature of an object is a measure of the thermal energy content 3.Chemical 3.Chemical — results from the particular arrangement of atoms in a chemical compound; radiant and thermal energy produced in this reaction due to energy released during the breaking and reforming of chemical bonds 4.Nuclear 4.Nuclear — radiant and thermal energy released when particles in the nucleus of the atoms are rearranged 5.Electrical 5.Electrical — due to the flow of electrically charged particles Energy & Chemistry

3. Potential & Kinetic Energy Kinetic energy  Potential energy  Potential energy – Energy stored in an object because of the relative positions or orientations of its components – PE = Fd = mad = mgh = work  Kinetic energy  Kinetic energy – Energy due to the motion of an object – KE = ½ mv 2

4. Potential & Kinetic Energy

5. Internal Energy (E) PE + KE = Internal energy (E or U) Internal Energy of a chemical system depends on number of particles type of particles temperature The higher the T the higher the internal energy So, use changes in T (∆T) to monitor changes in E (∆E).

6. heat transfer out (exothermic), -q heat transfer in (endothermic), +q SYSTEM ∆E = q + w w transfer in (+w) w transfer out (-w) Internal Energy (E)

7. Energy & Chemistry All of thermodynamics depends on the law of CONSERVATION OF ENERGY. The total energy is unchanged in a chemical reaction. If PE of products is less than reactants, the difference must be released as KE. Energy Change in Chemical Processes Potential Energy of system dropped. Kinetic energy increased. Therefore, you often feel a Temperature increase.

8. Thermodynamics -Enthalpy ThermodynamicsThermodynamics – the science of heat (energy) transfer. Heat transfers until thermal equilibrium is established. ∆T measures energy transferred. SYSTEM – The object under study 1. Open system — can exchange both matter and energy with its surroundings 2. Closed system — can exchange energy but not matter with its surroundings 3. Isolated system — exchanges neither energy nor matter with the surroundings; total energy of the system plus the surroundings is constant SURROUNDINGS – Everything outside the system

9. FIRST LAW OF THERMODYNAMICS ∆E = q + w heat energy transferred energychange work done by the system Energy is conserved!

10. The First Law of Thermodynamics Exothermic reactions generate specific amounts of heat. This is because the potential energies of the products are lower than the potential energies of the reactants.

11. The First Law of Thermodynamics There are two basic ideas of importance for thermodynamic systems. 1.Chemical systems tend toward a state of minimum potential energy. 2.Chemical systems tend toward a state of maximum disorder. The first law is also known as the Law of Conservation of Energy. –Energy is neither created nor destroyed in chemical reactions and physical changes.

12. The state of the system is a complete description of the system at a given time, including its temperature and pressure, the amount of matter it contains, its chemical composition, and the physical state of the matter Some examples of state functions are: –T (temperature), P (pressure), V (volume),  E (change in energy),  H (change in enthalpy – the transfer of heat), and S (entropy) Examples of non-state functions are: –n (moles), q (heat), w (work)  ∆H along one path =  ∆H along another path This equation is valid because ∆H is a STATE FUNCTION These depend only on the state of the system and not how it got there. V, T, P, energy — and your bank account! Unlike V, T, and P, one cannot measure absolute H. Can only measure ∆H. State of a System

13. Standard States and Standard Enthalpy Changes Thermochemical standard state conditions –The thermochemical standard T = 298.15 K. –The thermochemical standard P = 1.0000 atm. Be careful not to confuse these values with STP. Thermochemical standard states of matter –For pure substances in their liquid or solid phase the standard state is the pure liquid or solid. –For gases the standard state is the gas at 1.00 atm of pressure. For gaseous mixtures the partial pressure must be 1.00 atm. –For aqueous solutions the standard state is 1.00 M concentration. ∆H f o = standard molar enthalpy of formation the enthalpy change when 1 mol of compound is formed from elements under standard conditions.

14. State Function The properties of a system that depend only on the state of the system are called state functions. –State functions are always written using capital letters. The value of a state function is independent of pathway. An analog to a state function is the energy required to climb a mountain taking two different paths. –E 1 = energy on the first floor of Heldenfels –E 1 = mgh 1 –E 2 = energy on the fourth floor of Heldenfels –E 2 = mgh 2 –  E = E 2 -E 1 = mgh 2 – mgh 1 = mg(  h)

15. ENTHALPY Most chemical reactions occur at constant P, so and so ∆E = ∆H + w (and w is usually small) ∆H = heat transferred at constant P ≈ ∆E ∆H = change in heat content of the system ∆H = H final - H initial Heat transferred at constant P = q p q p = ∆H where H = enthalpy

16. Directionality of Heat Transfer Heat always transfer from hotter object to cooler one. EXO thermic: heat transfers from SYSTEM to SURROUNDINGS. T(system) goes down T(surr) goes up

17. Heat always transfers from hotter object to cooler one. ENDO thermic: heat transfers from SURROUNDINGS to the SYSTEM. T(system) goes up T (surr) goes down Directionality of Heat Transfer

18. If H final < H initial then ∆H is negative Process is EXOTHERMIC If H final > H initial then ∆H is positive Process is ENDOTHERMIC ENTHALPY ∆H = H final - H initial

19. Consider the formation of water H 2 (g) + 1/2 O 2 (g) → H 2 O(g) + 241.8 kJ USING ENTHALPY Exothermic reaction — heat is a “product” and ∆H = – 241.8 kJ

20. Example of HESS’S LAW— Making liquid H 2 O from H 2 + O 2 involves two exothermic steps. USING ENTHALPY Liquid H 2 O H 2 + O 2 gasH 2 O vapor Making liquid H 2 O from H 2 involves two steps. H 2(g) + 1/2 O 2(g) → H 2 O (g) + 242 kJ H 2 O (g) → H 2 O (l) + 44 kJ H 2 O (g) → H 2 O (l) + 44 kJ H 2(g) + 1/2 O 2(g) → H 2 O (l) + 286 kJ If a rxn. is the sum of 2 or more others, the net ∆H is the sum of the ∆H’s of the other rxns.

21. Enthalpy Values H 2(g) + 1/2 O 2(g) → H 2 O (g) ∆H˚ = -242 kJ 2H 2(g) + O 2(g) → 2H 2 O (g) ∆H˚ = -484 kJ H 2 O (g) → H 2(g) + 1/2 O 2(g) ∆H˚ = +242 kJ H 2(g) + 1/2 O 2(g) → H 2 O (l) ∆H˚ = -286 kJ Depend on how the reaction is written and on phases of reactants and products

22. Standard Molar Enthalpies of Formation,  H f o The standard molar enthalpy of formation is defined as the enthalpy for the reaction in which one mole of a substance is formed from its constituent elements. –The symbol for standard molar enthalpy of formation is  H f o. The standard molar enthalpy of formation for MgCl 2 is:

23. Standard molar enthalpies of formation have been determined for many substances and are tabulated in Table 15-1 and Appendix K in the text. Standard molar enthalpies of elements in their most stable forms at 298.15 K and 1.000 atm are zero. Example 15-4: The standard molar enthalpy of formation for phosphoric acid is -1281 kJ/mol. Write the equation for the reaction for which  H o rxn = - 1281 kJ. P in standard state is P 4 Phosphoric acid in standard state is H 3 PO 4(s) Standard Molar Enthalpies of Formation,  H f o

24. ½˚ H 2(g) + ½ O 2(g) → H 2 O (g) ∆H f ˚ (H 2 O, g )= -241.8 kJ/mol C (s) + ½ O 2(g) → CO (g) ∆H f ˚ of CO = - 111 kJ/mol By definition, ∆H f o = 0 for elements in their standard states. Use ∆H˚’s to calculate enthalpy change for H 2 O(g) + C(graphite) → H 2 (g) + CO(g) Standard Molar Enthalpies of Formation,  H f o

25. Calculate the enthalpy change for the reaction of one mole of H 2(g) with one mole of F 2(g) to form two moles of HF (g) at 25 o C and one atmosphere.

26. Standard Molar Enthalpies of Formation,  H f o Calculate the enthalpy change for the reaction in which 15.0 g of aluminum reacts with oxygen to form Al 2 O 3 at 25 o C and one atmosphere.

27.  Standard enthalpies of reaction (ΔH ° xn ) – The enthalpy that occurs when a reaction is carried out with all reactants and products in their standard states – General reaction: aA + bB  cC + dD where A, B, C, and D are chemical substances and a, b, c, and d are their stoichiometric coefficients – Magnitude of ΔH° rxn is the sum of the standard enthalpies of formation of the products, each multiplied by its appropriate coefficient, minus the sum of the standard enthalpies of formation of the reactants, multiplied by their coefficients ΔH° rxn = [cΔH° f (C) + dΔH° f (D)]  [aΔH° f (A) + bΔH° f (B)] or ΔH° rxn =  mΔH° f(product)   nΔH° f(reactants) where  means “sum of” and m and n are the stoichiometric coefficients of each of the products and reactants, respectively Hess’s Law – Enthalpies of Formation and Reaction

28. Hess’s Law & Energy Level Diagrams Forming H 2 O can occur in a single step or in a two steps. ∆H total is the same no matter which path is followed.

29.  Standard enthalpies of formation – The magnitude of ΔH for a reaction depends on the physical states of the reactants and products, the pressure of any gases present, and the temperature at which the reaction is carried out. – A specific set of conditions under which enthalpy changes are measured are used to ensure uniformity of reaction conditions and data. – Standard conditions: a pressure of 1 atmosphere (atm) for gases and a concentration of 1 M for species in solution (1 mol/L), each pure substance must be in its standard state, its most stable form at a pressure of 1 atm at a specified temperature. – Enthalpies of formation measured under standard conditions are called standard enthalpies of formation (ΔH ° f ). – Standard enthalpy of formation of any element in its standard state is zero. Enthalpies of Formation

30.  Hess’s law allows the calculation of the enthalpy change for any conceivable chemical reaction by using a relatively small set of tabulated data 1.Enthalpy of combustion, ΔH comb —enthalpy change that occurs when a substance is burned in excess oxygen 2.Enthalpy of fusion, ΔH fus —enthalpy change that accompanies the melting, or fusion, of 1 mol of a substance 3.Enthalpy of vaporization, ΔH vap —the enthalpy change that accompanies the vaporization of 1 mol of a substance 4.Enthalpy of solution, ΔH soln —enthalpy change when a specified amount of solute dissolves in a given quantity of solvent 5.Enthalpy of formation, ΔH f —enthalpy change for the formation of 1 mol of a compound from its component elements Enthalpies of Reaction

31. Hess’s Law Hess’s Law of Heat Summation,  H rxn =  H 1 +  H 2 +  H 3 +..., states that the enthalpy change for a reaction is the same whether it occurs by one step or by any (hypothetical) series of steps. –Hess’s Law is true because  H is a state function. If we know the following  H o ’s

32. For example, we can calculate the  H o for reaction [1] by properly adding (or subtracting) the  H o ’s for reactions [2] and [3]. Notice that reaction [1] has FeO and O 2 as reactants and Fe 2 O 3 as a product. –Arrange reactions [2] and [3] so that they also have FeO and O 2 as reactants and Fe 2 O 3 as a product. Each reaction can be doubled, tripled, or multiplied by half, etc. The  H o values are also doubled, tripled, etc. If a reaction is reversed the sign of the  H o is changed. Hess’s Law

33. Given the following equations and  H o  values calculate  H o for the reaction below. Hess’s Law

34. Calculate the  H o 298 for the following reaction: Hess’s Law

35. Application of Hess’s Law and more algebra allows us to calculate the  H f o  for a substance participating in a reaction for which we know  H rxn o, if we also know  H f o  for all other substances in the reaction. Given the following information, calculate  H f o for H 2 S (g).

36. HEAT CAPACITY The heat required to raise an object’s T by 1 ˚C. Which has the larger heat capacity? Thermal energy cannot be measured easily. Temperature change caused by the flow of thermal energy between objects or substances can be measured. Calorimetry is the set of techniques employed to measure enthalpy changes in chemical processes using calorimeters.

37. Specific Heat Capacity How much energy is transferred due to Temperature difference? The heat (q) “lost” or “gained” is related to a)sample mass b) change in T and c) specific heat capacity Specific heat capacity= heat lost or gained by substance (J) (mass, g) (T change, K)

38. Table of specific heat capacities SubstancePhasec p J g -1 K -1 C p J mol -1 K -1 Air (typ. room conditions) gas1.01229.19 Aluminiumsolid0.89724.2 Argongas0.520320.7862 Coppersolid0.38524.47 Diamondsolid0.50916.115 Ethanolliquid2.44112 Goldsolid0.129125.42 Graphitesolid0.7108.53 Heliumgas5.193220.7862 Hydrogengas14.3028.82 Ironsolid0.45025.1 Lithiumsolid3.5824.8 Mercuryliquid0.139527.98 Nitrogengas1.04029.12 Neongas1.030120.7862 Oxygengas0.91829.38 Uraniumsolid0.11627.7 Water gas (100 °C)2.08037.47 liquid (25 °C)4.181375.327 solid (0 °C)2.11438.09 All measurements are at 25 °C unless noted. Notable minimums and maximums are shown in maroon text. Aluminum  Heat capacity of an object depends on both its mass and its composition.  Molar heat capacity (Cp) — amount of energy needed to increase the temperature of 1 mol of a substance by 1°C; units of Cp are J/(mol°C)  Specific heat (Cs) — amount of energy needed to increase the temperature of 1 g of a substance by 1°C, units are J/(g°C)  The quantity of a substance, the amount of heat transferred, its heat capacity, and the temperature change are related in two ways: q = nCpΔT where n = number of moles of substance q = mCsΔT where m = mass of substance in grams

39. Specific Heat Capacity If 25.0 g of Al cool from 310 o C to 37 o C, how many joules of heat energy are lost by the Al?

40. Heat/Energy Transfer No Change in State q transferred = (sp. ht.)(mass)(∆T)

41. Heat Transfer Use heat transfer as a way to find specific heat capacity, C p 55.0 g Fe at 99.8 ˚C Drop into 225 g water at 21.0 ˚C Water and metal come to 23.1 ˚C What is the specific heat capacity of the metal?

42. Heating/Cooling Curve for Water Note that T is constant as ice melts or water boils

43. Thermochemical Equations Thermochemical equations are a balanced chemical reaction plus the  H value for the reaction. –For example, this is a thermochemical equation. The stoichiometric coefficients in thermochemical equations must be interpreted as numbers of moles. 1 mol of C 5 H 12 reacts with 8 mol of O 2 to produce 5 mol of CO 2, 6 mol of H 2 O, and releasing 3523 kJ is referred to as one mole of reactions.

44. This is an equivalent method of writing thermochemical equations.  H < 0 designates an exothermic reaction.  H > 0 designates an endothermic reaction Thermochemical Equations

45. Write the thermochemical equation for the reaction for CuSO 4(aq) + 2NaOH (aq) Cu(OH) 2(s) + Na 2 SO 4(aq) 50.0mL of 0.400 M CuSO4 at 23.35 o C T final 25.23 o C 50.0mL of 0.600 M NaOH at 23.35 o C Density final solution = 1.02 g/mL C H 2 O = 4.184 J/g o C Thermochemical Equations

46. Standard Temperature and Pressure Standard temperature and pressure is given the symbol STP. –It is a reference point for some gas calculations. Standard P  1.00000 atm or 101.3 kPa Standard T  273.15 K or 0.00 o C –Gas laws must use the Kelvin scale to be correct. Relationship between Kelvin and centigrade. At the macroscopic level, a complete physical description of a sample of a gas requires four quantities: 1. Temperature (expressed in K) 2. Volume (expressed in liters) 3. Amount (expressed in moles) 4. Pressure (given in atmospheres) These variables are not independent — if the values of any three of these quantities are known, the fourth can be calculated. Gas Laws

47. Boyle’s Law: The Volume-Pressure Relationship V  1/P or V= k (1/P) or PV = k P 1 V 1 = k 1 for one sample of a gas. P 2 V 2 = k 2 for a second sample of a gas. k 1 = k 2 for the same sample of a gas at the same T. Mathematically we write Boyle’s Law as P 1 V 1 = P 2 V 2 This relationship between pressure and volume is known as Boyle’s law which states that at constant temperature, the volume of a fixed amount of a gas is inversely proportional to its pressure. Robert Boyle (1627-1691). Son of Earl of Cork, Ireland.

48. Charles’ Law: The Volume-Temperature Relationship; The Absolute Temperature Scale Jacques Charles (1746-1823). Isolated boron and studied gases. Balloonist. Charles’s and Gay-Lussac’s findings can be stated as: At constant pressure, the volume of a fixed amount of a gas is directly proportional to its absolute temperature (in K). This relationship is referred to as Charles’s law and is stated mathematically as V = (constant) [T (in K)] or V  T (in K, at constant P).

49. Charles’ Law: The Volume-Temperature Relationship; The Absolute Temperature Scale Gay-Lussac showed that a plot of V versus T was a straight line that could be extrapolated to – 273.15ºC at zero volume, a theoretical state. The slope of the plot of V versus T varies for the same gas at different pressures, but the intercept remains constant at –273.15ºC. Significance of the invariant T intercept in plots of V versus T was recognized by Thomson (Lord Kelvin), who postulated that –273.15ºC was the lowest possible temperature that could theoretically be achieved, and he called it absolute zero (0 K).

50. The Combined Gas Law Equation Boyle’s and Charles’ Laws combined into one statement is called the combined gas law equation. –Useful when the V, T, and P of a gas are changing.

51. Avogadro’s Law and the Standard Molar Volume Avogadro’s Law states that at the same temperature and pressure, equal volumes of two gases contain the same number of molecules (or moles) of gas. standard molar volumeIf we set the temperature and pressure for any gas to be STP, then one mole of that gas has a volume called the standard molar volume. Stated mathematically: V = (constant) (n) or V  n (at constant T and P) The standard molar volume is 22.4 L at STP. –This is another way to measure moles. –For gases, the volume is proportional to the number of moles. 11.2 L of a gas at STP = 0.500 mole 44.8 L of a gas at STP = ? Moles

52. Boyle’s Law: The Volume-Pressure Relationship At 25 o C a sample of He has a volume of 4.00 x 10 2 mL under a pressure of 7.60 x 10 2 torr. What volume would it occupy under a pressure of 2.00 atm at the same T?

53. Charles’ Law: The Volume-Temperature Relationship; The Absolute Temperature Scale A sample of hydrogen, H 2, occupies 1.00 x 10 2 mL at 25.0 o C and 1.00 atm. What volume would it occupy at 50.0 o C under the same pressure?

54. The Combined Gas Law Equation A sample of nitrogen gas, N 2, occupies 7.50 x 10 2 mL at 75.0 0 C under a pressure of 8.10 x 10 2 torr. What volume would it occupy at STP?

55. Avogadro’s Law and the Standard Molar Volume One mole of a gas occupies 36.5 L and its density is 1.36 g/L at a given temperature and pressure. (a) What is its molar mass? (b) What is its density at STP?

56. Summary of Gas Laws: The Ideal Gas Law What volume would 50.0 g of ethane, C 2 H 6, occupy at 1.40 x 10 2 o C under a pressure of 1.82 x 10 3 torr?

57. Summary of Gas Laws: The Ideal Gas Law Calculate the number of moles in, and the mass of, an 8.96 L sample of methane, CH 4, measured at standard conditions?

58. Dalton’s Law of Partial Pressures John Dalton 1766-1844 1.The ideal gas law assumes that all gases behave identically. If V, T, and n for each gas in a mixture are known, the pressure of each gas, its partial pressure, can be calculated. Holding V and T constant, P is directly proportional to n: P = n(RT/V) = n(constant) 2.Generally, for a mixture of i components, the total pressure is given by P t = (n 1 + n 2 + n 3 + - - - +n i ) (RT/V). 3.The above equation makes it clear that, at constant T and V, P depends on only the total number of moles of gas present, whether the gas is a single chemical species or a mixture of gaseous species. Nothing in the equation depends on the nature of the gas, only on the quantity. 4. The total pressure exerted by a mixture of gases is the sum of the partial pressures of component gases. This law is known as Dalton’s law of partial pressures and can be written mathematically as P t = P 1 + P 2 + P 3 - - - + P i where P t is the total pressure and the other terms are the partial pressures of the individual gases. Vapor Pressure Vapor Pressure is the pressure exerted by a substance’s vapor over the substance’s liquid at equilibrium. Partial pressure Partial pressure is the pressure the gas would exert if it were the only one present (at the same temperature and volume).

59. Dalton’s Law of Partial Pressures If 1.00 x 10 2 mL of hydrogen, measured at 25.0 o C and 3.00 atm pressure, and 1.00 x 10 2 mL of oxygen, measured at 25.0 o C and 2.00 atm pressure, were forced into one of the containers at 25.0 o C, what would be the pressure of the mixture of gases?

60. Dalton’s Law of Partial Pressures A sample of hydrogen was collected by displacement of water at 25.0 o C. The atmospheric pressure was 748 torr. What pressure would the dry hydrogen exert in the same container?

61. Mass-Volume Relationships in Reactions Involving Gases 2 mol KClO 3 yields 2 mol KCl and 3 mol O 2 2(122.6g) yields 2 (74.6g) and 3 (32.0g) Those 3 moles of O 2 can also be thought of as: 3(22.4L) or 67.2 L at STP In this section we are looking at reaction stoichiometry, like in Chapter 3, just including gases in the calculations.

62. Mass-Volume Relationships in Reactions Involving Gases What volume of oxygen measured at STP, can be produced by the thermal decomposition of 120.0 g of KClO 3 ?

63. Real Gases: Deviations from Ideality J. van der Waals, 1837-1923, Professor of Physics, Amsterdam. Nobel Prize 1910. The Ideal Gas Law ignores both the volume occupied by the molecules of a gas and all interactions between molecules, whether attractive or repulsive. In reality, all gases have a volume and the molecules of real gases interact with one another. For an ideal gas, a plot of PV/nRT versus P gives a horizontal line with an intercept of 1 on the PV/nRT axis. Real gases behave ideally at ordinary temperatures and pressures. At low temperatures and high pressures real gases do not behave ideally. The reasons for the deviations from ideality are: The molecules are very close to one another, thus their volume is important. The molecular interactions also become important.

64. Real Gases: Deviations from Ideality The van der Waals’ equation (P + an 2 /V 2 ) (V – nb) = nRT low temperatures high pressures. accounts for the behavior of real gases at low temperatures and high pressures. At large volumes a and b are relatively small and the van der Waal’s equation reduces to the ideal gas law at high temperatures and low pressures. The van der Waals constants a and b are empirical constants that differ for each gas that take into account two things: Pressure term, P + (an 2 /V 2 ) a corrects for intermolecular attractive forces that tend to reduce the pressure from that predicted by the ideal gas law For nonpolar gases the attractive forces are London Forces For polar gases the attractive forces are dipole-dipole attractions or hydrogen bonds. n 2 /V 2 represents the concentration of the gas (n/V) squared because it takes two particles to engage in the pairwise intermolecular interactions Volume term, V – nb, corrects for the volume occupied by the gaseous molecules b accounts for volume of gas molecules

65. Real Gases: Deviations from Ideality Calculate the pressure exerted by 84.0 g of ammonia, NH 3, in a 5.00 L container at 200. o C using the ideal gas law. PV = nRT P = nRT/V n = 84.0g * 1mol/17 g T = 200 + 273 P = (4.94mol)(0.08206 L atm mol -1 K -1 )(473 K) (5 L) P = 38.3 atm

66. Real Gases: Deviations from Ideality Calculate the pressure exerted by 84.0 g of ammonia, NH 3, in a 5.00 L container at 200. o C using the van der Waal’s equation. The van der Waal's constants for ammonia are: a = 4.17 atm L 2 mol -2 b = 3.71x10 -2 L mol -1 n = 84.0g * 1mol/17 g T = 200 + 273 P = (4.94mol)(0.08206 L atm mol -1 K -1 )(473K) (4.94 mol) 2 *4.17 atm L 2 mol -2 5 L – (4.94 mol*3.71 E -2 L mol -1 ) (5 L) 2 P = 39.81 atm – 4.07 atm = 35.74 P = 38.3 atm 7% error