Question

A scientist measures the standard enthalpy change for the following reaction to be 81.1 kJ :2CO2(g) + 5 H2(g)C2H2(g) + 4 H2O(g)Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of CO2(g) is _____

Answer 1

Answer: The enthalpy of the formation of $CO_2(g)$ is coming out to be -410.8 kJ/mol.Z

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:  

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

For the given chemical reaction:

$2CO_2(g)+5H_2(g)\rightarrow C_2H_2(g)+4H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(C_2H_2(g))})+(4\times \Delta H^o_f_{(H_2O(g))})]-[(2\times \Delta H^o_f_{(CO_2(g))})+(5\times \Delta H^o_f_{(H_2(g))})]$

We are given:

$\Delta H^o_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H^o_f_{(H_2(g))}=0kJ/mol\\\Delta H^o_f_{(C_2H_2(g))}=226.7kJ/mol\\\Delta H^o_{rxn}=81.1kJ$

Putting values in above equation, we get:

$81.1=[(1\times (226.7)})+(4\times (-241.8))]-[(2\times \Delta H^o_f_{(CO_2(g))})+(5\times (0))]\\\\\Delta H^o_f_{(CO_2(g))}=-410.8kJ/mol$

Hence, the enthalpy of the formation of $CO_2(g)$ is coming out to be -410.8 kJ/mol.