Question

Top fuel dragsters and funny cars burn nitro-methane as fuel according to the following balanced combustion equation: 2CH3NO2(l)+3/2O2(g)→2CO2(g)+3H2O(g)+N2(g). The standard enthalpy of combustion for nitromethane is −709.2kJ/mol. Calculate the standard enthalpy of formation(delta h formation) for nitro-methane.

Answer 1

Answer : The standard enthalpy of formation for nitro-methane is, -467.4 kJ

Explanation :

Enthalpy change : It 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:  

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

The equilibrium reaction follows:

$2CH_3NO_2(l)+\frac{3}{2}O_2(g)\rightleftharpoons 2CO_2(s)+3H_2O(g)+N_2(g)$

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

$\Delta H^o_{rxn}=[(n_{(CO_2)}\times \Delta H^o_f_{(CO_2)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})+(n_{(N_2)}\times \Delta H^o_f_{(N_2)})]-[(n_{(CH_3NO_2)}\times \Delta H^o_f_{(CH_3NO_2)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]$

We are given:

$\Delta H^o_{rxn}=-709.2kJ/mol$

$\Delta H^o_f_{(CH_3NO_2(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-286kJ/mol\\\Delta H^o_f_{(N_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$-709.2=[(2\times -393)+(3\times -286)+(1\times 0)]-[(2\times \Delta H^o_f_{(CH_3NO_2)})+(\frac{3}{2}\times 0)]$

$\Delta H^o_f_{(CH_3NO_2)}=-467.4kJ$

Thus, the standard enthalpy of formation for nitro-methane is, -467.4 kJ