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stepan [7]
3 years ago
5

The heat of combustion per mole for acetylene, C2H2(g), is -1299.5 kJ/mol. Assuming that the combustion products are CO2(g) and

H2O(l), and given that the enthalpy of formation is -393.5 kJ/mol for CO2(g) and -285.8 kJ/mol for H2O(l), find the enthalpy of formation of C2H2(g).
Chemistry
1 answer:
guapka [62]3 years ago
3 0

Answer: The heat of combustion per mole for acetylene is 227.7 kJ/mol.

Explanation:

The combustion equation of acetylene is as follows.

C_{2}H_{2} + \frac{5}{2}O_{2} \rightarrow 2CO_{2} + H_{2}O

Formula to calculate enthalpy of formation for a reaction is as follows.

\Delta H^{o}_{rxn} = \sum \Delta H_{products} - \sum \Delta H_{reactants}\\\Delta H^{o}_{rxn} = [2\Delta H^{o}_{f}(CO_{2}) + \Delta H^{o}_{f} (H_{2}O)] - [\Delta H^{o}_{f}(C_{2}H_{2}) + \frac{5}{2} \Delta H^{o}_{f} O_{2}]\\-1299.5 = 2(-393.5) + (-285.8) - \Delta H^{o}_{f} (C_{2}H_{2})\\\Delta H^{o}_{f} (C_{2}H_{2}) = 227.7 kJ/mol

Thus, we can conclude that heat of combustion per mole for acetylene is 227.7 kJ/mol.

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