Chapter 6 Notes Thermoche mistry

Part 1: Energy, Heat and Work Thermoche mistry

Energy is the capacity to do work or to produce heat. The law of conservation of energy states that energy can be converted from one form to another, but can be neither created nor destroyed.

Kinetic energy is energy due to the motion of the object and depends on the mass and velocity of the object. KE = ½mv 2 Potential energy is energy due to position or composition. **Mass must be in kilograms and velocity must be in meters/second!!!!! The unit kg  m 2 = J s 2

Heat and temperature are different. Temperature is a measure of the kinetic energy of the molecules. Heat refers to the transfer of energy between two objects due to a temperature difference. Heat is not a substance contained by an object, although we often talk of heat as if this were true.

The pathway is likened to a “path” or “route.” For instance, I can get to the stadium by walking out the front door or the back door. A state function or state property depends only on the characteristics of the present state – not on the pathway.

The universe is divided into two parts: a. The system is the part of the universe on which we focus. b. The surroundings include everything else in the universe. For a reaction, the system includes the reactants and products. The surroundings includes the reaction container, the room, etc. (i.e. anything else other than reactants and products.)

The study of energy and its interconversions is called thermodynamics. The law of conservation of energy is often called the first law of thermodynamics. It states: The energy of the universe is constant.

Thermodynamic quantities always consist of two parts: a number, giving the magnitude of the change, and a sign, indicating the direction of the flow. The sign reflects the system’s point of view.

q = heat *If energy flows INTO the system via heat (endothermic), then q = +. *If energy flows OUT OF the system via heat (exothermic), then q = .

w = work *If the surroundings do work on the system (energy flows into the system), then w = +. * If the system does work on the surroundings (energy flows out of the system), then w = .

Work is defined as force acting over a distance. The formula used to solve for work is: W = PΔV Pressure is measured in atmospheres and volume is measured in liters. Convert to joules using J = 1 L  atm The sign changes depending on: compressed gas = +PΔV expanding gas =  PΔV

Memorize!!! vaporization melting sublimation condensation freezing deposition gas endothermicliquidexothermic solid

The internal energy, E, of a system is defined as the sum of the kinetic and potential energies. The formula is ΔE = q + w where q is heat and w is work. The sign convention is that anything that leaves the system is negative. *q is negative: system releases heat *q is positive: system absorbs heat *w is negative: system does work *w is positive: surroundings do work

Part 2: Propertie s of Enthalpy Thermoche mistry

Enthalpy, H, is defined as: H = E + PV where E = internal energy P = pressure V = volume

Internal energy, pressure and volume are all state functions (independent of the pathway) therefore enthalpy is also a state function.

At constant pressure, the change in enthalpy (ΔH) of the system is equal to the energy flow as heat. Therefore, ΔH = q at constant P. At constant pressure, exothermic means that ΔH is negative and endothermic means that ΔH is positive.

Stoichiometric Calculations When a mole of methane (CH 4 ) is burned at constant pressure, 890 kJ of energy are released as heat. Calculate  H when 5.8 grams of methane are burned at constant pressure g/mol CH O 2  CO H 2 O  H = –890 kJ 5.8 g x kJ – 890. kJ1 mol CH g CH 4 = – 320 kJ 1 mol CH g CH 4

Part 3: Calorimetr y and Heat Capacity Thermoche mistry

Calorimetry is the study of heat flow and heat measurement. Calorimetry

Calorimetry experiments determine the heats (enthalpy changes) of reactions by making accurate measurements of temperature changes produced in a calorimeter.

The formula used in calorimetric calculations is: q = mc  T where q = heat (J) m = mass (g) c = specific heat capacity (J/g  C)  T = change in temperature (  C)

The heat capacity of an object is the amount of heat needed to raise the temperature of the object by 1  C. The heat capacity of one gram of a substance is called its specific heat.

The specific heat is a physical property of the substance, like its color and melting point. Substances have different specific heat capacities.

The specific heat capacity of water = 4.18 J/g  C

When calculating  T, always subtract the smaller temperature FROM the larger temperature:  T = T larger  T smaller **If the temperature rises, (ex: from 25  C to 30  C) then q will be negative and the reaction is exothermic. **If the temperature drops, (ex: from 40  C to 30  C) then q will be positive and the reaction is endothermic.

Example 1: What is the specific heat capacity of iron if the temperature of a 12.3-g sample of iron is increased by 10.2  C when 56.7 J of heat is added?

q = mc  T Example 1: What is the specific heat capacity of iron if the temperature of a 12.3-g sample of iron is increased by 10.2  C when 56.7 J of heat is added? 56.7 J = (12.3 g)( c)(10.2  C) 56.7 J=(12.3 g)(c) (10.2  C) J/g  C = c (12.3 g)(10.2  C)

Example 2: When a 13.7-g sample of solid Pb(NO 3 ) 2 dissolves in 85.0 g of water in a calorimeter, the temperature drops from 23.4  C to 19.7  C. Calculate  H for the solution process. Pb(NO 3 ) 2 (s)  Pb +2 (aq) + 2 NO 3  1 (aq)  H = ?

q = mc  T Example 2: When a 13.7-g sample of solid Pb(NO 3 ) 2 dissolves in 85.0 g of water in a calorimeter, the temperature drops from 23.4  C to 19.7  C. Calculate  H for the solution process. Pb(NO 3 ) 2 (s)  Pb +2 (aq) + 2 NO 3  1 (aq)  H = ? q= (85.0 g) (4.18 J/g  C ) (23.4  C  19.7  C ) q = 1310 J **Since the temperature dropped, q will be positive and the reaction is endothermic. q = mc(T larger  T smaller )

Example 2: When a 13.7-g sample of solid Pb(NO 3 ) 2 dissolves in 85.0 g of water in a calorimeter, the temperature drops from 23.4  C to 19.7  C. Calculate  H for the solution process. Pb(NO 3 ) 2 (s)  Pb +2 (aq) + 2 NO 3  1 (aq)  H = ? Calculate the molar mass of Pb(NO 3 ) 2 : MM = 331 g/mol Remember that from the first part, q = J

Example 2: When a 13.7-g sample of solid Pb(NO 3 ) 2 dissolves in 85.0 g of water in a calorimeter, the temperature drops from 23.4  C to 19.7  C. Calculate  H for the solution process. Pb(NO 3 ) 2 (s)  Pb +2 (aq) + 2 NO 3  1 (aq)  H = ? J 13.7 g Pb(NO 3 ) g Pb(NO 3 ) 2 1 mol Pb(NO 3 ) 2 = +31,700 J = 31.7 kJ

Energy Calculations

Important Information: q = mcΔT q = ΔH fusion  moles q = ΔH vaporization  moles specific heat capacity of water = 4.18 J/g  C specific heat capacity of ice = 2.1 J/g  C specific heat capacity of steam = 1.8 J/g  C ΔH fusion of water = 6.0 kJ/mole ΔH vaporization of water = 40.7 kJ/mole

How much energy does it take to convert 130. grams of ice at  40.0  C to steam at 160.  C?

Convert grams to moles of water: 1 mol H 2 O130. g H 2 O 18.0 g H 2 O = 7.22 mol H 2 O

Plan:a. Heat ice from  40.0  C to 0.00  C.

q = mc  T q = (130. g)( 2.1 J/g  C)(40.0  C) q = 10,920 J q  10,900 J q = 10.9 kJ

b. Add heat to convert ice to liquid water at 0  C.

q =  H fusion  moles q = (6.0 kJ/mol)(7.22 mol) q = kJ q  43.3 kJ

c. Heat liquid water from 0.00  C to 100.  C.

q = mc  T q = (130. g)( 4.18 J/g  C)(100.0  C) q = 54,340 J q  54,300 J q = 54.3 kJ

d. Add heat to convert liquid water to steam at 100  C.

q =  H vaporization  moles q = (40.7 kJ/mol)(7.22 mol) q = kJ q  294 kJ

e. Heat steam from 100.  C to 160.  C.

q = mc  T q = (130. g)( 1.8 J/g  C)(60.0  C) q = 14,040 J q  14.0 kJ

Add the energy values: Total energy = a + b + c + d + e

Total energy = 10.9 kJ 43.3 kJ 54.3 kJ 294 kJ 14.0 kJ kJ  417 kJ

Part 4: Hess’ Law Thermoche mistry

The amount of heat that a reaction absorbs or releases depends on the conditions under which the reaction is carried out (temperature, pressure, and physical states of the reactants and products.)

To make comparing enthalpy changes easier, chemists chose a pressure of 1 atmosphere and a temperature of 25  C as conditions to carry out reactions. These are called standard states. An enthalpy change under these conditions is called a standard enthalpy change. It is denoted with a superscript. It is shown as  H .

Conventional Definitions of Standard States For a compound: *The standard state of a gaseous substance is a pressure of exactly 1 atmosphere. *For a pure substance in a condensed state (liquid or solid), the standard state is the pure liquid or solid. *For a substance present in a solution, the standard state is a concentration of exactly 1 M. For an element: *The standard state of an element is the form in which the element exists under conditions of 1 atmosphere and 25  C. (The standard state for oxygen is O 2 (g) at a pressure of 1 atm; the standard state for sodium is Na(s); the standard state for mercury is Hg(ℓ).

In the 19 th century, a Swiss chemist named G.H. Hess proposed a way of finding the enthalpy change for a reaction (even if the reaction could not be performed directly.) In 1840, Hess demonstrated experimentally that the heat transferred during a given reaction is the same whether the reaction occurs in one step or several steps.

His method is now called Hess’ law of heat summation or simply Hess’ Law. The method is analogous to solving simultaneous equations in algebra.

In solving problems using Hess’ Law, there are some basic rules that must be memorized.

Rules for manipulating reactions: 1. If a reaction is reversed, the sign of  H must be reversed. 2. If a reaction is multiplied or divided by a coefficient,  H must also be multiplied or divided by that coefficient.

Rules for adding reactions: 1. Identical substances on the same side of a reaction are added together. 2. Identical substances on opposite sides of a reaction are cancelled. 3. Simply add the  H’s of each reaction to get the  H of the final reaction.

Use: A + B  C + D  H =  10 kJ 2 E + C  D + 2 F  H =  20 kJ E + A  F  H =  30 kJ Determine  H for 3 A + B  2C Example 1:

A + B  C + D  H =  10 kJ 2E + C  D + 2F  H =  20 kJ E + A  F  H =  30 kJ Determine  H for 3A + B  2C A + B  C + D  H =  10 kJ 2 F + D  2 E + C  H = + 20 kJ E + A  F  H =  30 kJ 222 2( ) Write the reactions by comparing the substances with the desired result.

A + B  C + D  H =  10 kJ 2E + C  D + 2F  H =  20 kJ E + A  F  H =  30 kJ Determine  H for 3A + B  2C A + B  C + D  H =  10 kJ 2 F + D  2 E + C  H = + 20 kJ E + A  F  H =  30 kJ) 2222( Cancel identical items on opposite sides of the arrow.

A + B  C + D  H =  10 kJ 2E + C  D + 2F  H =  20 kJ E + A  F  H =  30 kJ Determine  H for 3A + B  2C A + B  C + D  H =  10 kJ 2 F + D  2 E + C  H = + 20 kJ E + A  F  H =  30 kJ) 2222( Add identical items on the same side of the arrow. 3 A2C+B 

A + B  C + D  H =  10 kJ 2E + C  D + 2F  H =  20 kJ E + A  F  H =  30 kJ Determine  H for 3A + B  2C A + B  C + D  H =  10 kJ 2 F + D  2 E + C  H = + 20 kJ E + A  F  H =  30 kJ) 2222( Verify that the final reaction matches the reaction given in the original problem. 3 A + B  2C

A + B  C + D  H =  10 kJ 2E + C  D + 2F  H =  20 kJ E + A  F  H =  30 kJ Determine  H for 3A + B  2C A + B  C + D  H =  10 kJ 2 F + D  2 E + C  H = + 20 kJ E + A  F  H =  30 kJ) 2222( Combine the  H’s. 3 A2C+B  H =  50 kJ

When dealing with actual substances in the reactions, the physical state of each substance must be written in parentheses. For example: (s) is solid, (ℓ) is liquid, (g) is gas, and (aq) is aqueous – meaning that it is dissolved in water.

From the following enthalpy changes, 2PbS (s) + 3 O 2 (g)  2 PbO (s) + 2 SO 2 (g)  H  =  124 kJ Pb (s) + CO (g)  PbO (s) + C (s)  H  =  kJ calculate the value of  H for the following reaction: 2PbS (s) + 3 O 2 (g) + 2 C (s)  2 Pb (s) + 2 CO (g) + 2 SO 2 (g) Example 2:

2PbS (s) + 3 O 2 (g)  2 PbO (s) + 2 SO 2 (g)  H  =  124 kJ Find  H for: 2PbS (s) + 3 O 2 (g) + 2 C (s)  2 Pb (s) + 2 CO (g) + 2 SO 2 (g) Using: 2PbS (s) + 3 O 2 (g)  2 PbO (s) + 2 SO 2 (g)  H  =  124 kJ Pb (s) + CO (g)  PbO (s) + C (s)  H  =  kJ Leave the first reaction as written:

2PbS (s) + 3 O 2 (g)  2 PbO (s) + 2 SO 2 (g)  H  =  124 kJ Find  H for: 2PbS (s) + 3 O 2 (g) + 2 C (s)  2 Pb (s) + 2 CO (g) + 2 SO 2 (g) PbO (s) + C (s)  Pb (s) + CO (g)  H  = kJ Using: 2PbS (s) + 3 O 2 (g)  2 PbO (s) + 2 SO 2 (g)  H  =  124 kJ Pb (s) + CO (g)  PbO (s) + C (s)  H  =  kJ Reverse the second reaction and change the sign:

2PbS (s) + 3 O 2 (g)  2 PbO (s) + 2 SO 2 (g)  H  =  124 kJ Find  H for: 2PbS (s) + 3 O 2 (g) + 2 C (s)  2 Pb (s) + 2 CO (g) + 2 SO 2 (g) PbO (s) + C (s)  Pb (s) + CO (g)  H  = kJ) Using: 2PbS (s) + 3 O 2 (g)  2 PbO (s) + 2 SO 2 (g)  H  =  124 kJ Pb (s) + CO (g)  PbO (s) + C (s)  H  =  kJ Multiply the second reaction by 2: 22222(

2PbS (s) + 3 O 2 (g)  2 PbO (s) + 2 SO 2 (g)  H  =  124 kJ Find  H for: 2PbS (s) + 3 O 2 (g) + 2 C (s)  2 Pb (s) + 2 CO (g) + 2 SO 2 (g) PbO (s) + C (s)  Pb (s) + CO (g)  H  = kJ) Using: 2PbS (s) + 3 O 2 (g)  2 PbO (s) + 2 SO 2 (g)  H  =  124 kJ Pb (s) + CO (g)  PbO (s) + C (s)  H  =  kJ Cancel identical items on opposite sides of the arrow (

2PbS (s) + 3 O 2 (g)  2 PbO (s) + 2 SO 2 (g)  H  =  124 kJ Find  H for: 2PbS (s) + 3 O 2 (g) + 2 C (s)  2 Pb (s) + 2 CO (g) + 2 SO 2 (g) PbO (s) + C (s)  Pb (s) + CO (g)  H  = kJ) Using: 2PbS (s) + 3 O 2 (g)  2 PbO (s) + 2 SO 2 (g)  H  =  124 kJ Pb (s) + CO (g)  PbO (s) + C (s)  H  =  kJ 22222( 2PbS (s) + 3 O 2 (g) + 2 C (s)  2 Pb (s) + 2 CO (g) + 2 SO 2 (g) Add identical items on the same side of the arrow.

2PbS (s) + 3 O 2 (g)  2 PbO (s) + 2 SO 2 (g)  H  =  124 kJ Find  H for: 2PbS (s) + 3 O 2 (g) + 2 C (s)  2 Pb (s) + 2 CO (g) + 2 SO 2 (g) PbO (s) + C (s)  Pb (s) + CO (g)  H  = kJ) Using: 2PbS (s) + 3 O 2 (g)  2 PbO (s) + 2 SO 2 (g)  H  =  124 kJ Pb (s) + CO (g)  PbO (s) + C (s)  H  =  kJ Verify that the final reaction matches the reaction given in the original problem ( 2PbS (s) + 3 O 2 (g) + 2 C (s)  2 Pb (s) + 2 CO (g) + 2 SO 2 (g)

2PbS (s) + 3 O 2 (g)  2 PbO (s) + 2 SO 2 (g)  H  =  124 kJ Find  H for: 2PbS (s) + 3 O 2 (g) + 2 C (s)  2 Pb (s) + 2 CO (g) + 2 SO 2 (g) PbO (s) + C (s)  Pb (s) + CO (g)  H  = kJ) Using: 2PbS (s) + 3 O 2 (g)  2 PbO (s) + 2 SO 2 (g)  H  =  124 kJ Pb (s) + CO (g)  PbO (s) + C (s)  H  =  kJ 22222( 2PbS (s) + 3 O 2 (g) + 2 C (s)  2 Pb (s) + 2 CO (g) + 2 SO 2 (g) Combine the  H’s.  H  = kJ

From the following enthalpy changes, C (graphite) + O 2 (g)  CO 2 (g)  H  =  394 kJ H 2 (g) + ½O 2 (g)  H 2 O (ℓ)  H  =  286 kJ CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (ℓ)  H  =  kJ calculate the value of  H for the following reaction: C (graphite) + 2 H 2 (g)  CH 4 (g) Example 3:

C (graphite) + O 2 (g)  CO 2 (g)  H  =  394 kJ Find  H for: C (graphite) + 2 H 2 (g)  CH 4 (g) Using: C (graphite) + O 2 (g)  CO 2 (g)  H  =  394 kJ H 2 (g) + ½O 2 (g)  H 2 O (ℓ)  H  =  286 kJ CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (ℓ)  H  =  kJ Leave the first reaction as written:

C (graphite) + O 2 (g)  CO 2 (g)  H  =  394 kJ H 2 (g) + ½O 2 (g)  H 2 O (ℓ)  H  =  286 kJ Leave the second reaction as written: Find  H for: C (graphite) + 2 H 2 (g)  CH 4 (g) Using: C (graphite) + O 2 (g)  CO 2 (g)  H  =  394 kJ H 2 (g) + ½O 2 (g)  H 2 O (ℓ)  H  =  286 kJ CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (ℓ)  H  =  kJ

C (graphite) + O 2 (g)  CO 2 (g)  H  =  394 kJ H 2 (g) + ½O 2 (g)  H 2 O (ℓ)  H  =  286 kJ Reverse the third reaction and change the sign: Find  H for: C (graphite) + 2 H 2 (g)  CH 4 (g) Using: C (graphite) + O 2 (g)  CO 2 (g)  H  =  394 kJ H 2 (g) + ½O 2 (g)  H 2 O (ℓ)  H  =  286 kJ CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (ℓ)  H  =  kJ CO 2 (g) + 2 H 2 O (ℓ)  CH 4 (g) + 2 O 2 (g)  H  = kJ

C (graphite) + O 2 (g)  CO 2 (g)  H  =  394 kJ H 2 (g) + ½O 2 (g)  H 2 O (ℓ)  H  =  286 kJ) Multiply the second reaction by 2: 2 22 22( Find  H for: C (graphite) + 2 H 2 (g)  CH 4 (g) Using: C (graphite) + O 2 (g)  CO 2 (g)  H  =  394 kJ H 2 (g) + ½O 2 (g)  H 2 O (ℓ)  H  =  286 kJ CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (ℓ)  H  =  kJ CO 2 (g) + 2 H 2 O (ℓ)  CH 4 (g) + 2 O 2 (g)  H  = kJ

C (graphite) + O 2 (g)  CO 2 (g)  H  =  394 kJ H 2 (g) + ½O 2 (g)  H 2 O (ℓ)  H  =  286 kJ) Cancel identical items on opposite sides of the arrow. 2 22 22( Find  H for: C (graphite) + 2 H 2 (g)  CH 4 (g) Using: C (graphite) + O 2 (g)  CO 2 (g)  H  =  394 kJ H 2 (g) + ½O 2 (g)  H 2 O (ℓ)  H  =  286 kJ CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (ℓ)  H  =  kJ CO 2 (g) + 2 H 2 O (ℓ)  CH 4 (g) + 2 O 2 (g)  H  = kJ

C (graphite) + O 2 (g)  CO 2 (g)  H  =  394 kJ H 2 (g) + ½O 2 (g)  H 2 O (ℓ)  H  =  286 kJ) Add identical items on the same side of the arrow. 2 22 22( Find  H for: C (graphite) + 2 H 2 (g)  CH 4 (g) Using: C (graphite) + O 2 (g)  CO 2 (g)  H  =  394 kJ H 2 (g) + ½O 2 (g)  H 2 O (ℓ)  H  =  286 kJ CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (ℓ)  H  =  kJ CO 2 (g) + 2 H 2 O (ℓ)  CH 4 (g) + 2 O 2 (g)  H  = kJ C (graphite) + 2 H 2 (g)  CH 4 (g)

C (graphite) + O 2 (g)  CO 2 (g)  H  =  394 kJ H 2 (g) + ½O 2 (g)  H 2 O (ℓ)  H  =  286 kJ) 2 22 2 2( Find  H for: C (graphite) + 2 H 2 (g)  CH 4 (g) Using: C (graphite) + O 2 (g)  CO 2 (g)  H  =  394 kJ H 2 (g) + ½O 2 (g)  H 2 O (ℓ)  H  =  286 kJ CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (ℓ)  H  =  kJ CO 2 (g) + 2 H 2 O (ℓ)  CH 4 (g) + 2 O 2 (g)  H  = kJ C (graphite) + 2 H 2 (g)  CH 4 (g) Verify that the final reaction matches the reaction given in the original problem.

C (graphite) + O 2 (g)  CO 2 (g)  H  =  394 kJ H 2 (g) + ½O 2 (g)  H 2 O (ℓ)  H  =  286 kJ) Combine the  H’s. 2 22 22( Find  H for: C (graphite) + 2 H 2 (g)  CH 4 (g) Using: C (graphite) + O 2 (g)  CO 2 (g)  H  =  394 kJ H 2 (g) + ½O 2 (g)  H 2 O (ℓ)  H  =  286 kJ CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (ℓ)  H  =  kJ CO 2 (g) + 2 H 2 O (ℓ)  CH 4 (g) + 2 O 2 (g)  H  = kJ C (graphite) + 2 H 2 (g)  CH 4 (g)  H  =  75.7 kJ

Part 5: Standard Enthalpies of Formation Thermoche mistry

The standard enthalpy of formation (ΔH f  ) of a compound is defined as the change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states. The ΔH f  values for some common substances are shown in Table 6.2. More values are found in Appendix 4. **ΔH f  for an element in its standard state is zero.

The change in enthalpy for a given reaction can be calculated from the enthalpies of formation of the reactants and products: ΔH˚rxn = Σ n p ΔH f ˚ products  Σ n r ΔH f ˚ reactants

Use the standard enthalpies of formation in Appendix 4 to calculate the standard enthalpy change for the overall reaction that occurs when ammonia is burned in air to form nitrogen dioxide and water. 4 NH 3 (g) + 7 O 2 (g) → 4 NO 2 (g) + 6 H 2 O (ℓ) ΔH˚ rxn = Σ n p ΔH f ˚ products  Σ n r ΔH f ˚ reactants ΔH˚ rxn = [(4 mol NO 2 ) (34 kJ/mol) + (6 mol H 2 O) (  286 kJ/mol)]  [(4 mol NH 3 ) (  46 kJ/mol) + (7 mol O 2 ) (0 kJ/mol)] ΔH˚ rxn = [(136 kJ) + (  1716 kJ)]  [(  184 kJ) + (0 kJ)] ΔH˚ rxn = [  1580 kJ]  [  184 kJ] ΔH˚ rxn =  1396 kJ

It is to your benefit to work every assigned homework problem because you won’t be able to memorize how to work all of the problems. The test over this chapter will consist of a multiple choice section (75%) and an essay section (25%). The test will be given over a period of two days. This format is similar to the AP Chemistry exam. On the AP exam, each section is worth 50%.