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Standard XII

Chemistry

Law of Mass Action

N2 and H2 a...

Question

$N_{2}$ and $H_{2}$ are taken in $1 : 3$ molar ratio in a closed vessel to attain the following equilibrium $N_{2} g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ . Find $k_{p}$ for reaction at total pressure of $2 P$ if $P_{N_{2}}$ at equilibrium is $\frac{P}{3}$ .

\frac{1}{3 P^{2}}

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\frac{4}{3 P^{2}}

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\frac{4 P^{2}}{3}

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Solution

The correct option is B $\frac{4}{3 P^{2}}$
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)$
$\frac{P}{3}$ $3 \times \frac{P}{3}$ $2 \times \frac{P}{3}$

$∴ K_{p} = \frac{p_{{N H}_{3}}^{2}}{p_{N_{2}} \times p_{H_{2}}^{3}} = \frac{{(2 \times \frac{P}{3})}^{2}}{(\frac{P}{3}) \times {(3 \times \frac{P}{3})}^{3}} = \frac{4 P^{2} / 9}{P^{4} / 3} = \frac{4}{3 P^{2}}$

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Similar questions

N_{2}

and

H_{2}

are taken in 1 : 3 molar ratio in a closed vessel to attained the following equilibrium

N_{2 (g)} + 3 H_{2 (g)} ⇌ 2 N H_{3 (g)}

Find

K_{p}

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P_{N_{2}}

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1

mole

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and

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N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)

. The equilibrium constant

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{N H}_{3}

is:

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

for the reaction initially the mole ratio was

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Q. 1 mole

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N2 and H2 are taken in 1:3 molar ratio in a closed vessel to attain the following equilibrium N2g)+3H2(g)⇌2NH3(g). Find kp for reaction at total pressure of 2P if PN2 at equilibrium is P3.

The correct option is B 43P2N2(g)+3H2(g)⇌2NH3(g)P3 3×P3 2×P3∴Kp=p2NH3pN2×p3H2=(2×P3)2(P3)×(3×P3)3=4P2/9P4/3=43P2

$N_{2}$ and $H_{2}$ are taken in $1 : 3$ molar ratio in a closed vessel to attain the following equilibrium $N_{2} g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ . Find $k_{p}$ for reaction at total pressure of $2 P$ if $P_{N_{2}}$ at equilibrium is $\frac{P}{3}$ .