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Byju's Answer

Standard XII

Chemistry

Characteristics of Chemical Equilibrium

At 1400 K, ...

Question

At $1400 K, K_{c} = 2.5 \times 10^{- 3}$ for reaction $C H_{4} (g) + 2 H_{2} S (g) ⇌ C S_{2} (g) + 4 H_{2} (g)$ . At 10.0L reaction vessel at 1400K contains 2.00 mole of CH $_{4}$ , 3.0 mol of CS $_{2}$ , 3.0 mole of H $_{2}$ and 4.0 mole of H $_{2}$ S. Then ________.

this reaction is at equilibrium with above concentrations.

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the reaction will proceed in forward direction to reach equilibrium.

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the reaction will proceed in backward direction to each equilibrium.

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the information is insufficient to decide the direction of progress of reaction.

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Solution

The correct option is A the reaction will proceed in backward direction to each equilibrium.
$Q_{c} = (\frac{{(\frac{3}{10})}^{4} (\frac{3}{10})}{(\frac{2}{10}) {(\frac{4}{10})}^{2}} = \frac{243}{32} \times 10^{- 2} = 7.59 \times 10^{- 2} > K_{c}$
So, reaction will proceed in backward direction.

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Similar questions

Q. At

1400 K

, for the reaction

C H_{4} (g) + 2 H_{2} S (g) ⇌ C S_{2} (g) + 4 H_{2} (g)

K_{c} = 2.5 \times 10^{- 3}

. A

10 L

reaction vessel contains 2 moles of

C H_{4},

3 moles of

C S_{2}

, 3 moles of

H_{2}

and 4 moles of

H_{2} S

. Choose the correct statement from the following:

Q. At

1400 K

K_{c} = 2.5 \times 10^{- 3}

for the reaction

C H_{4} (g) + 2 H_{2} S (g) ⇌ C S_{2} (g) + 4 H_{2} (g)

. If A

10.0 L

reaction vessel at

1400 K

contains

2.0

moles of CH

_{4}

3.0

moles of

C S_{2}

3.0

moles of

H_{2}

and

4.0

moles of

H_{2} S

, then:

Q. At 1400 K,

K_{c} = 2.5 \times 10^{3}

for the reaction

C H_{4} (g) + 2 H_{2} S ⇌ C S_{2} (g) + 4 H_{2} (g)

. A 10.0 L reaction vessel at 1400 K contains 2.0 mol of

C H_{4}

, 3.0 mol of

C S_{2}

, 3.0 mol of

H_{2}

and 4.0 mol of

H_{2} S

.The reaction will proceed from right to left to reach equilibrium.

Q. The equilibrium constant for the following reaction is 49 at

450^{o} C .

0.22 m o l

I_{2}

0.22 m o l

H_{2}

and

0.66 m o l

H I

were put in a evacuated 1 litre container.

H_{2} + I_{2} ⇌ 2 H I

Then, the correct statement will be:

Q. Equilibrium constant,

K_{C}

for the reaction

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

500 K

0.061

At a particular time, the analysis shows that composition of the reaction mixture is

3.0 m o l L^{- 1} N_{2}, 2.0 m o l L^{- 1} H_{2}

and

0.5 m o l L^{- 1} N H_{3} .

Is the reaction at equilibirium and if not in which direction does the reaction tend to proceed to reach equilibrium?

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At 1400K,Kc=2.5×10−3 for reaction CH4(g)+2H2S(g)⇌CS2(g)+4H2(g). At 10.0L reaction vessel at 1400K contains 2.00 mole of CH4, 3.0 mol of CS2, 3.0 mole of H2 and 4.0 mole of H2S. Then ________.

The correct option is A the reaction will proceed in backward direction to each equilibrium.Qc=((310)4(310)(210)(410)2=24332×10−2=7.59×10−2>KcSo, reaction will proceed in backward direction.

At $1400 K, K_{c} = 2.5 \times 10^{- 3}$ for reaction $C H_{4} (g) + 2 H_{2} S (g) ⇌ C S_{2} (g) + 4 H_{2} (g)$ . At 10.0L reaction vessel at 1400K contains 2.00 mole of CH $_{4}$ , 3.0 mol of CS $_{2}$ , 3.0 mole of H $_{2}$ and 4.0 mole of H $_{2}$ S. Then ________.

The correct option is A the reaction will proceed in backward direction to each equilibrium.
$Q_{c} = (\frac{{(\frac{3}{10})}^{4} (\frac{3}{10})}{(\frac{2}{10}) {(\frac{4}{10})}^{2}} = \frac{243}{32} \times 10^{- 2} = 7.59 \times 10^{- 2} > K_{c}$
So, reaction will proceed in backward direction.