Q. When $NaNO_{3}$ is heated in a closed, oxygen is liberated and $NaNO_{2}$ is is left behind. At equilibrium

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addition of $NaNO_{2}$ favours forward reverse reaction

addition of $NaNO_{2}$ favours forward reaction

increasing temperature favours forward reaction

decreasing pressure favours reverse reaction

Solution:

$NaNO_{3}\left(\right.s\left.\right)\rightleftharpoonsNaNO_{2}\left(\right.s\left.\right)+\frac{1}{2}O_{2}\left(\right.g\left.\right);\Delta H=+ve$
Also $K_{p}=\left(\right.P_{O_{2}}\left(\left.\right)^{1 / 2}$ and thus, addition of $NaNO_{3}$ or $NaNO_{2}$ has no effect.

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