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Grade upto college level Physical Chemistry

Give the following standard heats of reaction :
(i) heat of formation of water = -68.3 kcal;
(ii) heat of combustion of acetylene = - 310.6 kcal;
(iii) heat of combustion of ethylene = -337.2 kcal;
Calculate the heat of reaction for the hydrogenation of acetylene at constant volume (25oC).

Profile image of Shane Macguire
12 Years agoGrade upto college level
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1 Answer

Profile image of Deepak Patra
12 Years ago
The given data can be written as follows
(i) H2(g) + 1/2 O2(g) → H2O(l); ∆ = - 68.3 kcal
(ii) C2H2(g) + 5/2 O2(g) → H2O(l) + 2CO2(g); ∆H = - 310.6 kcal
(iii) C2H4(g) + 3O2 → 2H2O(l) + 2CO2(g); ∆H = - 337.2kcal
The required thermochemical equation is
C2H2(g) + H2(g) → C2H4(g)
The required equation can be obtained by subtracting equation (iii) from the sum of equations (i) and (ii), thus ∆H of the required equation can be calculated as below.
∆H = [-68.3 + (-310.6)] – (-337.2)
= [-68.3 – 310.6] + 337.2
= - 378.9 + 337.2 = - 41.7 kcal
∆E = ∆H - ∆nRT
Here ∆n = Moles of the gaseous products – Moles of the gaseous reactants
= 1 – (1 + 1) = - 1
Substituting the value of ∆H, ∆n, R and T in
∆E = ∆H - ∆nRT
∆E = - 41.7 – (- 1 * 2 * 10-3 * 298)
[∵ R = 2cal/ degree/ mole = 2 * 10-3 kcal/ deg/ mole]
= - 41.7 + 2 * 10-3 * 298
= - 41.7 + 0.596 = 41.104 kcal

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