When NaNO_3 is heated in a closed vessel,O_2 | vedclass

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(D) The chemical reaction is: $2NaNO_3(s) \rightleftharpoons 2NaNO_2(s) + O_2(g)$.According to Le Chatelier's principle:$1$. Adding more reactant $(Na

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All of the above

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(D) The chemical reaction is: $2NaNO_3(s) ightleftharpoons 2NaNO_2(s) + O_2(g)$.According to Le Chatelier's principle:$1$. Adding more reactant $(NaNO_3)$ shifts the equilibrium in the forward direction to consume the excess reactant.$2$. Adding more product $(NaNO_2)$ shifts the equilibrium in the reverse direction to consume the excess product.$3$. Increasing the pressure in a system with gaseous products shifts the equilibrium toward the side with fewer moles of gas. Here,the product side has $1 	ext{ mole}$ of $O_2(g)$ and the reactant side has $0 	ext{ moles}$ of gas. Thus,increasing pressure shifts the equilibrium to the left (reverse reaction).Therefore,all the given statements are correct.

When $NaNO_3$ is heated in a closed vessel,$O_2$ is liberated and $NaNO_2$ is left behind. At equilibrium,

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