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Effect on Kc

At 1400 K, ...

Question

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:

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 reach 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 D the reaction will proceed in backward direction to reach equilibrium
The number of moles of $C H_{4}, H_{2} S, C S_{2}$ and $H_{2}$ are 2.0, 4.0, 3.0 and 3.0 respectively.

The volume is 10 L.

The molar concentrations of $C H_{4}, H_{2} S, C S_{2}$ and $H_{2}$ are 0.2M, 0.4M, 0.3M and 0.3M respectively.

The expression for the reaction quotient is $Q = \frac{[C S_{2}] [H_{2}]^{4}}{[C H_{4}] [H_{2} S]^{2}} = \frac{0.3 \times (0.3)^{4}}{0.2 \times (0.4)^{2}} = 0.0759$ .

This is much greater than the value of the equilibrium constant $K_{c} = 2.5 \times 10^{- 3}$ .

Hence, the reaction will proceed in backward direction to reach equilibrium.

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Similar questions

Q. At

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At 1400K, Kc=2.5×10−3 for the reaction CH4(g)+2H2S(g)⇌CS2(g)+4H2(g). If A 10.0L reaction vessel at 1400K contains 2.0 moles of CH4, 3.0 moles of CS2, 3.0 moles of H2 and 4.0 moles of H2S, then:

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: