Question

Consider the reaction:

$4N{O}_2\left(g\right)+O_2\left(g\right)\to 2N_2{O}_5\left(g\right),{\Delta }_{r}H=-111kJ$. If $N_2{O}_5\left(s\right)$ is formed instead of $N_2{O}_5\left(g\right)$ in the above reaction, the ${\Delta }_{r}H$ value will be:

(given, $\Delta H$ of sublimation for $N_2{O}_5$ is $54kJmo{l}^{-1}$)

• A. + 54 kJ
• B. + 219 kJ
• C. − 219 J
• D. −165 kJ

Answer 1

The correct option is D −165 kJ

${\Delta }_{r}H=-111kJ$

$-111-54=\Delta H^{\prime }$

$\Delta H^{\prime }=-165kJ$