Question

For a particular isomer of C8H18, the combustion reaction produces 5113.3 kJ of heat per mole of C8H18(g) consumed, under standard conditions. C8H18(g)+252O2(g)⟶8CO2(g)+9H2O(g)ΔH∘rxn=−5113.3 kJ/mol What is the standard enthalpy of formation of this isomer of C8H18(g)? ΔH∘f=

Answer 1

Answer: The standard enthalpy of formation of this isomer of $C_8H_{18}(g)$ is -210.9 kJ

Explanation:

The given balanced chemical reaction is,

$C_8H_{18}(g)+\frac{25}{2}O_2(g)\rightarrow 8CO_2(g)+9H_2O(g)$

First we have to calculate the enthalpy of formation of  $C_8H_{18}$.

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{CO_2}\times \Delta H_f^0_{(CO_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{O_2}\times \Delta H_f^0_{(O_2)+n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})}]$

where,

We are given:

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

Putting values in above equation, we get:

$-511.3kJ/mol=[(8\times -393.5)+(9\times -241.8)]-[(\frac{25}{2}\times 0)+(1\times \Delta H_f^0_{(C_8H_{18})}$

$\Delta H_f^0_{(C_8H_{18})}=-210.9kJ$

Answer 2

Answer: The enthalpy of formation of $C_8H_{18}(g)$ is -210.98 kJ/mol

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.

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

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

For the given chemical reaction:

$C_8H_{18}(g)+\frac{25}{2}O_2(g)\rightarrow 8CO_2(g)+9H_2O(g)$

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

$\Delta H_{rxn}=[(8\times \Delta H_f_{(CO_2(g))})+(9\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(C_8H_{18}(g))})+(\frac{25}{2}\times \Delta H_f_{(O_2(g))})]$

We are given:

$\Delta H_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.51kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol\\\Delta H_{rxn}=-5113.3kJ$

Putting values in above equation, we get:

$-5113.3=[(8\times (-393.51))+(9\times (-241.8))]-[(1\times \Delta H_f_{(C_8H_{18}(g))})+(\frac{25}{2}\times (0))]\\\\\Delta H_f_{(C_8H_{18}(g))}=-210.98kJ/mol$

Hence, the enthalpy of formation of $C_8H_{18}(g)$ is -210.98 kJ/mol