For the reaction 2NO(g) + O_2(g) arrow 2NO_2(g) at 400 K , the...

The given reaction is $2\mathrm{NO}(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightarrow 2\mathrm{NO}_2(\mathrm{g})$

The change in the number of moles of gaseous species is:

$\Delta n_g = n_p - n_r = 2 - (2 + 1) = -1$

The relation between enthalpy change and internal energy change is:

$\Delta H = \Delta U + \Delta n_g RT$

Substituting the given values:

$\Delta H = -70 + (-1) \times (8.314 \times 10^{-3}) \times 400$

$\Delta H = -70 - 3.3256$

$\Delta H = -73.3256\text{ kJ} \approx -73.33\text{ kJ}$

Answer: $-73.33\text{ kJ}$