nitrogen and hydrogen react to form ammonia, like this: n₂(g)+3h₂(g)→2nh₃(g) suppose a mixture of n₂, h₂ and nh₃ has come to equilibrium in a closed reaction vessel. predict what change, if any, the perturbations in the table below will cause in the composition of the mixture in the vessel. also decide whether the equilibrium shifts to the right or left. perturbation change in composition shift in equilibrium some nh₃ is added. the pressure of n₂ will select to the right the pressure of h₂ will select to the left (none) some n₂ is removed. the pressure of h₂ will select to the right the pressure of nh₃ will select to the left (none)

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# Explanation:
## Step1: Le - Chatelier's principle for adding $NH_3$
When some $NH_3$ is added, according to Le - Chatelier's principle, the system will try to counteract the change. Since $NH_3$ is a product ($N_2(g)+3H_2(g)
ightleftharpoons2NH_3(g)$), increasing the concentration (or pressure, for gases in a closed vessel) of $NH_3$ will shift the equilibrium to the left (to consume the added $NH_3$).
As the equilibrium shifts to the left, more $N_2$ and $H_2$ are formed. So, the pressure of $N_2$ will increase (because the reverse reaction produces $N_2$) and the pressure of $H_2$ will increase (because the reverse reaction produces $H_2$ in a 1:3 ratio with $N_2$ for the reverse reaction).

## Step2: Le - Chatelier's principle for removing $N_2$
When some $N_2$ is removed, the system will try to replace the removed $N_2$. Since $N_2$ is a reactant ($N_2(g)+3H_2(g)
ightleftharpoons2NH_3(g)$), decreasing the concentration (or pressure) of $N_2$ will shift the equilibrium to the left (to produce more $N_2$).
As the equilibrium shifts to the left, $H_2$ is consumed (because the reverse reaction consumes $H_2$ in a 1:3 ratio with $N_2$ for the forward reaction) and $NH_3$ is consumed (because the reverse reaction consumes $NH_3$). So, the pressure of $H_2$ will increase (no, wait, wrong. Wait, when $N_2$ is removed, equilibrium shifts left. The reaction $2NH_3
ightarrow N_2 + 3H_2$. So, $H_2$ is produced as the reaction shifts left. Wait, no: original reaction $N_2+3H_2
ightarrow2NH_3$. If $N_2$ is removed, reverse reaction $2NH_3
ightarrow N_2 + 3H_2$ occurs. So, $H_2$ pressure increases (because it is a product of the reverse reaction) and $NH_3$ pressure decreases (because it is consumed in the reverse reaction)

# Answer:
|perturbation|change in composition|shift in equilibrium|
|---|---|---|
|Some $NH_3$ is added|The pressure of $N_2$ will increase; The pressure of $H_2$ will increase|to the left|
|Some $N_2$ is removed|The pressure of $H_2$ will increase; The pressure of $NH_3$ will decrease|to the left|

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Explanation:

Step1: Le - Chatelier's principle for adding \(NH_3\)

Step2: Le - Chatelier's principle for removing \(N_2\)

Answer: