At 450oC- the equilibrium constant Kp for the reaction- N 2 g-3 H 2 g- 2 NH 3 g- was found to be 1-6-10-5 at a pressure of 200 atm- If N2 and H2 are taken in 1-3 ratio- what is

#160; #160; #160; #160; #160; #160; #160; #160; N _ 2 (g) + 3 H _ 2 (g)⇌ #160; 2 NH _ 3 (g) At eq. #160; #160; # ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Characteristics of Equilibrium Constant

At 450oC, t...

Question

At $450^{o} C$ , the equilibrium constant $K_{p}$ for the reaction, $N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)$ , was found to be $1.6 \times 10^{- 5}$ at a pressure of $200 a t m$ . If $N_{2}$ and $H_{2}$ are taken in $1 : 3$ ratio, what is % of ${N H}_{3}$ formed at this temperature?

Open in App

Solution

$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)$
At eq. $(1 - x)$ $(3 - 3 x)$ $2 x$

Total number of moles $= 1 - x + 3 - 3 x + 2 x = 4 - 2 x$

$p_{N_{2}} = \frac{(1 - x)}{(4 - 2 x)} P; p_{H_{2}} = \frac{(3 - 3 x)}{(4 - 2 x)} P; p_{{N H}_{3}} = \frac{2 x}{(4 - 2 x)} P$

$K_{p} = \frac{{(p_{{N H}_{3}})}^{2}}{p_{N_{2}} \times {(p_{H_{2}})}^{2}} = \frac{4 x^{2} {(4 - 2 x)}^{2}}{(1 - x) \times 27 \times {(1 - x)}^{3} P^{2}}$

$.6 \times 10^{- 5} = \frac{16}{27} \times \frac{x^{2} {(2 - x)}^{2}}{{(1 - x)}^{4} \times 200^{2}}$

or, $x = 0.30$

Moles of ammonia formed $= 2 \times 0.30$

total moles at equilibrium $= (4 - 2 x) = (4 - 2 \times 0.30) = 3.40$

% of ${N H}_{3}$ at equilibrium $\frac{0.60}{3.40} \times 100 = 17.64$

Suggest Corrections

Similar questions

Q. At

45^{0} C

the equilibrium constant

K_{P}

for the reaction

N_{2} + {3 H}_{2} ⇌ {2 N H}_{3}

was found to be

1.6 \times 10^{- 5}

at a pressure of 200 atm. If

N_{2}

and

H_{2}

are taken in a ratio 1:3. What is the % of

{N H}_{3}

formed at this temperature?

Q. If a mixture of

N_{2}

and

H_{2}

in the ratio 1 : 3 at 50 atmosphere and

650^{o} C

is allowed to react till equilibrium is reached. Ammonia present at equilibrium was at 25 atm pressure. Calculate the equilibrium constant (

K_{P}

) for the reaction.

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

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

P_{N_{2}}

at equilibrium is

\frac{P}{3}

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}

for reaction, if total pressure at equilibrium is 2P and

P_{N_{2}}

is P/3.

1

mole

N_{2}

and

3

moles

H_{2}

are placed in a closed container at a pressure of

4

atm. The pressure falls to

3

atm at the same temperature when the equilibrium is attained

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

. The equilibrium constant

K_{p}

for dissociation of

{N H}_{3}

is:

Join BYJU'S Learning Program

At 450oC, the equilibrium constant Kp for the reaction, N2(g)+3H2(g)⇌2NH3(g), was found to be 1.6×10−5 at a pressure of 200 atm. If N2 and H2 are taken in 1:3 ratio, what is % of NH3 formed at this temperature?

At $450^{o} C$ , the equilibrium constant $K_{p}$ for the reaction, $N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)$ , was found to be $1.6 \times 10^{- 5}$ at a pressure of $200 a t m$ . If $N_{2}$ and $H_{2}$ are taken in $1 : 3$ ratio, what is % of ${N H}_{3}$ formed at this temperature?