When NaNO_{3(s)} is heated in a closed vessel | vedclass

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Question

(C) According to Le Chatelier's principle:
$1$. $NaNO_3$ and $NaNO_2$ are solids. Adding more solid does not change their active mass,so it does

Vedclass · Accepted Answer

Increase in pressure shifts the reaction in the backward direction.

Vedclass · Answer

(C) According to Le Chatelier's principle:
$1$. $NaNO_3$ and $NaNO_2$ are solids. Adding more solid does not change their active mass,so it does not shift the equilibrium.
$2$. The reaction involves the production of $O_{2(g)}$. Increasing the pressure will shift the equilibrium towards the side with fewer moles of gas,which is the backward direction (towards $NaNO_3$).
$3$. The decomposition of $NaNO_3$ is an endothermic process $(\Delta H > 0)$. According to Le Chatelier's principle,decreasing the temperature will shift the equilibrium in the backward direction (exothermic direction).
$4$. Therefore,option $C$ is correct because increasing pressure favors the side with fewer gaseous moles.

When $NaNO_{3(s)}$ is heated in a closed vessel,$O_{2(g)}$ is released and $NaNO_{2(s)}$ remains. The equilibrium is: $NaNO_{3(s)} \rightleftharpoons NaNO_{2(s)} + 1/2 O_{2(g)}$. Which of the following is correct?

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