Question

N2 + 3h2 = 2nh3 for the reaction initially the mole ratio was 1:3 of n2:h2 at equilibrium 50% of each has reacted if the equilibrium pressure is p the partial preaaure of nh3 at equilibrium will be ?

Answer 1

Answer : The partial pressure of $NH_3$ at equilibrium is $(\frac{p}{3})$.

Solution : Given,

Initial mole ratio of $N_2:H_2$ is 1 : 3

At equilibrium, 50% of $N_2,H_2$ has reacted.

Total pressure = p

The given balanced reaction is,

                            $N_2+3H_2\rightarrow 2NH_3$

initial moles          x       3x            0

At equilibrium, 50% of $N_2,H_2$ has reacted that means the moles will become half.

The moles of $N_2$ = $(x-\frac{x}{2})$

The moles of $H_2$ = $(3x-\frac{3x}{2})$

The moles of $NH_3$ = $\frac{2x}{2}$ = x

Total moles at equilibrium = $(x-\frac{x}{2})+(3x-\frac{3x}{2})+x=3x$

Now we have to calculate the partial pressure of the $NH_3$ at equilibrium.

Formula used :

$\text{ Partial pressure of }NH_3=\frac{\text{ Moles of }NH_3}{\text{ Total moles}}\times \text{ Total pressure}$

Now put all the given values in this formula, we get

$\text{ Partial pressure of }NH_3=\frac{x}{3x}\times p=\frac{p}{3}$

Therefore, the partial pressure of $NH_3$ at equilibrium is $(\frac{p}{3})$.

Answer 2

Explanation:

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