Using these data: (a) $N_2(g)$ $+$ $2O_2(g)$ $\longrightarrow$ $2NO_2(g)$ $\Delta H_a $ $= $ $67.7\; kJ$ (b) $N_2(g)$ $+$ $2O_2(g)$ $\longrightarrow$ $N_2O_4(g)$ $\Delta H_b = 9.7\; kJ$, calculate the standard enthalpy of the following reaction: (c) $2NO_2(g)$ $\longrightarrow$ $N_2O_4(g)$ $\Delta H_c $ $= $ $?$
Consider the following states: State 1: $2NO_2(g)$ State 2: $N_2O_4(g)$ State 3: $N_2(g)$ $+$ $2O_2(g)$ So we have: $\Delta H_a$ $ =$ $ \Delta H_{31}$ $\Delta H_b$ $=$ $\Delta H_{32}$ $\Delta H_c$ $=$ $\Delta H_{12}$
If we want to apply Hess's law: $\Delta H_{12}$ $=$ $\Delta H_{13}$ $+$ $\Delta H_{32}$, reaction (a) must be inversed:: (-a) $2NO_2(g) $ $\longrightarrow$ $N_2(g)+2O_2(g)$ $\Delta H_{-a}$ $ =$ $ \Delta H_{13}$ $=$ $\color{red}-\color{black}67.7 \;kJ$ (b) $N_2(g)+2O_2(g)$ $\longrightarrow$ $N_2O_4(g)$ $\Delta H_b$ $ =$ $\Delta H_{32}$ $=$ $9.7\; kJ$
Now we sum up the two equations and simplify applying Hess's law: (-a)+(b) $2NO_2(g)$ $ +$ $ N_2(g)$ $+$ $2O_2(g)$ $\longrightarrow$ $N_2(g)$ $+$ $2O_2(g)$ $+$ $N_2O_4(g)$ $\Delta H_{13}+\Delta H_{32}$ (-a)+(b)=(c) (c) $2NO_2(g)$ $\longrightarrow$ $N_2O_4(g)$ $\Delta H_{13}$ $+$ $\Delta H_{32}$ $ = $ $\Delta H_{12}(Hesse!)$ $ =$ $ -67.7$ $+$ $9.7$ $=$ $-58 \;kJ$