At a temperature of 298 K, the standard enthalpy formation of carbon dioxide, CO

Question

At a temperature of 298 K, the standard enthalpy formation of carbon dioxide, CO_2(g) is -393.5 kJ middot mol^-1, and the standard enthalpy of formation of water vapor, H_2O(g) is -241.8 kJ middot mol^-1. The standard enthalpy of reaction for the combustion of acetylene, C_2H_2(g) in oxygen at 298 K is: 2 C_2H_2(g) + 5 O_2(g) rightarrow 4 CO_2 (g) + 2 H_2O(g) is -2511.0 kJ. The standard enthalpy of acetylene, C_2 H_2(g), at 298 K, is: -453.4 kJ middot mol^-1 -226.7 kJ middot mol^-1 +113.4 kJ mol^-1 +226.7 kJ middot mol^-1 +453.4 kJ middot mol^-1 +906.8 kJ +1875.7 kJ middot mol^-1 None of the above; the correct standard enthalpy of formation of acetylene is _____ kJ middot mol^-1 Cannot tell, because the standard enthalpy of O_2(g) at 298 K is not given.

Explanation / Answer

Q12

HRxn = Hproducts - Hreactants

HRxn = (4*-393.5 + 2*-241.8) - (2*x+ 5*0)

we know that, the heat of combustion for this reaction

1081.2 kJ/mol

buet we have 2 moles, so 2*1081.2 = 2162.4

so..

HRxn = (4*-393.5 + 2*-241.8) - (2*x+ 5*0)

-2162.4= (4*-393.5 + 2*-241.8) - (2*x+ 5*0)

solving for x:

best aswer is fo C2H2 --> 52.4 kJ/mol (formation)

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