Question #141187

3. Given the following thermochemical equations,

C2H2(g) + 5/2 O2(g) → 2CO2(g) + H2O(l) ∆H = -1299.5 KJ

C(s) + O2(g) → CO2(g) ∆H = -393.5 KJ

H2(g) + ½ O2(g) → H2O(l) ∆H = -285.8 KJ

Calculate ∆H for the decomposition of one mole of acetylene, C2H2(g), to its elements in their stable state at 25oC and 1 atm.

C2H2(g) + 5/2 O2(g) → 2CO2(g) + H2O(l) ∆H = -1299.5 KJ

C(s) + O2(g) → CO2(g) ∆H = -393.5 KJ

H2(g) + ½ O2(g) → H2O(l) ∆H = -285.8 KJ

Calculate ∆H for the decomposition of one mole of acetylene, C2H2(g), to its elements in their stable state at 25oC and 1 atm.

Expert's answer

From the equations given;

"\\Delta_{H_a}=\\Delta_{H_b}+\\Delta_{H_c}"

The equation for the end reaction is;

"C_2H_2(-1299.5kJ)\\to 2CO_2(-393.5kJ)+H_2O(-285.8kJ)"

"\\Delta H\u00b0_D=\\Sigma \\Delta H\u00b0_{products}-\\Sigma \\Delta H\u00b0_{reactants}"

"=(-1299.5kJ)-[(2\\times -393.5kJ)+(-285.8kJ)]"

"=(-1299.5kJ)-(-1072.8kJ)"

"=-226.7kJ"

"\\Delta H\u00b0_D" For the reaction is therefore "-226.7kJ"

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