Question

the molar heats of combustion of C2H2 gas ,graphite and h2 gas are -310.62 ,-94.05 and -68.32 kcal respectively. the enthalpy of formation in f C2H2 gas is

Answer 1

Answer : The enthalpy of formation of $C_2H_2$ gas is -101.11 Kcal.

Solution : Given,

The enthalpy of reaction $C_2H_2$, $\Delta H^o_1$ = -310.62 Kcal

The enthalpy of reaction $C_{\text{ graphite}$, $\Delta H^o_2$ = -94.05 Kcal

The enthalpy of reaction $H_2$, $\Delta H^o_3$ = -68.32 Kcal

The balanced combustion reactions of $C_2H_2$, $C_{\text{ graphite}}$ and $H_2$ are,

$2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(l)$      ........(1)

$C_\text{ graphite}+O_2(g)\rightarrow CO_2(g)$       .........(2)

$H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$         ..........(3)

The formation reaction for $C_2H_2$ gas is,

$2C_\text{ graphite}+H_2(g)\rightarrow C_2H_2(g)$        ..........(4)

For the formation of $C_2H_2$ gas, reverse the equation(1) & divided by 2 and multiply equation(2) by 2 then adding all three equations, we get

$2CO_2(g)+H_2O(l)\rightarrow C_2H_2(g)+\frac{5}{2}O_2(g)$

$2C_\text{ graphite}+2O_2(g)\rightarrow 2CO_2(g)$

$H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$

The net reaction : $2C_\text{ graphite}+H_2(g)\rightarrow C_2H_2(g)$

The calculation for enthalpy of formation of $C_2H_2$ by using Hess's law:

$\Delta H_f(C_2H_2)=\frac{\Delta H^o_1}{2}+ 2\times \Delta H^o_2+\Delta H^o_3$

Now put all the given values in this formula, we get

$\Delta H_f(C_2H_2)=\frac{310.62Kcal}{2}+ 2\times (-94.05Kcal)+(-68.32Kcal)$ = -101.11 Kcal