Inorganic chemistry Assistance Lecturer Amjad Ahmed Jumaa  Calculation of ∆H from ∆E.  Thermo chemical equations.  Calculating heat absorbed or released using specific heat data. 1

 Calculation of ∆H from ∆E: Let us consider a reaction: aA+ bB→ cC +dD  Change in number of moles: = no. of moles of products – no. of moles of reactants. = (c +d) – (a+b) =∆n We can do that by this relation shape (by this equation): ∆H =∆E+∆n ×RT.

 It may be pointed out that while determining the value of ∆H, only the number of moles of gaseous reactants and products are taken into consideration.  The temperature is a Kelvin, the value of gas constant(R), is taken either in calories or joules per degree per mol and is cal (approximately 2 calories), or joules. Solved problem: The heat of combustion of ethylene at 17ºC, and at constant volume is kcals. Calculate the heat of combustion at constant pressure considering water to be in liquid state. R value is equal 2 cal degree -1 mol -1 ?

Solution: The chemical equation for the combustion of ethylene is: C 2 H 4 (g) +3O 2 (g) → 2CO 2 (g) +2H 2 O (l) No. of moles of the products= 2 No. of mole of the reactants= 4 ∆n = (2-4) =-2 ∆H = ∆E+∆n×RT. ∆E = kcal. T = = 290k. R = 2cals = 2 ×10 -3 kcals. ∆H = ×10 -3 ×-2×290 = kcal.

 Thermo chemical equations:  Equations showing both the mass and enthalpy relations are called thermo chemical equation. 1-The stiochiometric coefficients always refer to the number of moles of each substance. 2-When an equation is reversed, the roles of reactants and products change. Consequently, the magnitude of (ΔH), for the equation remains the same, but its sign changes. 3-If both sides of a thermo chemical equation are multiplied by a factor (n), then(ΔH), must also be multiplied by the same factor.

4-When writing thermo chemical equations, the physical states of all reactants and products must be specified, because they help determine the actual enthalpy changes. Example: Given the thermo chemical equation: SO 2 (g ) + 1 /2 O 2 (g ) SO 3 (g ) ΔH = - 99 kJ How much heat is evolved when: 1 /2 mol of (SO 2 (g ) ) reacts. 3 mol of (SO 2 (g ) ) reacts.

Solution : Heat evolved = (1 / 2) (99 kJ) = 5.0 x 10 1 kJ. Following the same argument as in part a: Heat evolved = 3 (99 kJ) = 3.0 x 10 2 kJ.

 Why isn't there a negative sign in our answer? The sign convention for an exothermic reaction( energy, as heat, is released by the system) is negative (-ΔH).  However, in the above example, we state that heat is evolved, so a negative sign is unnecessary.  Tip Remember that the heat evolved or absorbed (q) by a reaction carried out under constant pressure conditions is equal to the enthalpy change of the system, (ΔH).

 Calculating heat absorbed or released using specific heat data:  The equation for calculating the heat change is given by: q = msΔt ………………. (1) or q = CΔt where (m) is the mass of the sample and (Δt) is the temperature change. Δt = t final – t initial

Example: How much heat is absorbed by (80.0 g) of iron (Fe) when its temperature is raised from (25 C°) to (500 C°) ? the specific heat of iron is (0.444 J / g. C°) Solution: Notice that the specific heat of iron is given in the problem. Therefore, we should Use the equation (1) to solve this problem. Step(1) : calculate (Δt). Δt = t final – t initial = (500 C°) -(25 C°) = 475 C°.

Step(2) : substitute the known values into equation (1). q = m Fe s Fe Δt. q = (80.0 g ) (0.444 J /g. C°) (475 C°) = 1.69 x 10 4 J.