Question

What is the Hf for this reaction? 16CO2(g) + 18H2O(g) 2C8H18(l) + 25O2(g)

Answer 1

Answer: The enthalpy of reaction is 500.4 kJ/mol

Explanation:

The chemical equation follows:

$16CO_2(g)+18H_2O(g)\rightarrow 2C_8H_18(l)+25O_2(g)$

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

$\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(C_8H_18(l))})+(25\times \Delta H^o_f_{(O_2(g))})]-[(16\times \Delta H^o_f_{(CO_2(g))})+(18\times \Delta H^o_f_{(H_2O(l))})]$

$\Delta H^o_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(C_8H_{18}(l))}=-5074kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(2\times (-5074))+(25\times (0))]-[(16\times (-393.5))+(18\times (-241.8))]\\\\\Delta H^o_{rxn}=500.4kJ/mol$

The enthalpy of reaction is 500.4 kJ/mol

Answer 2

Answer:

10,148 kJ

Explanation:

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