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At 1400 K, K( c) = 2.5 xx 10^(-3) for th...

At `1400 K, K_( c) = 2.5 xx 10^(-3)` for the reaction
`CH_(4)(g)+2H_(2)S(g)hArrCS_(2)(g)+4H_(2)(g)`
A ` 10 L` reaction vessel at `1400 K` contains `2.0 mol` of `CH_(4)`, `3.0 mol` of `CS_(2)`, `3.0 mol` of `H_(2)S`. In which direction does the reaction proceed to reach equilibrium?

A

Forward

B

Backward

C

May be forward or backward

D

Reaction is in equilibrium

Text Solution

Verified by Experts

The correct Answer is:
B
`CH_(4)(g)+2H_(2)S(g)overset(K)(hArr)CS_(2)(g)+4H_(2)(g)`
or `K=([CS_(2)][H_(2)]^(2))/([CH_(4)][H_(2)S]^(2)) ...(i), [CS_(2)]=(3/10),[H_(2)]=(3/10),[CH_(4)]=(2/10)` and `[H_(2)S]=(4/10)`
Substituting the cocentration values in equation (i).
`K=((3/10)(3/10)^(4))/((2/10)(4/10))=3^(5)/10^(5)xx10^(3)/(2xx4^(4))xx243/3200=0.075`
Therefore, `K gt 2.5xx10^(-3)` (given). Hence, the reaction will proceed in backward direction.
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Knowledge Check

  • At 1400 K, K_c=2.5xx10^(-3) for the reaction CH_4(g)+2H_2S(g)hArrCS_2(g)+4H_2(g) .A 10.0 L reaction vessel at 1400 K contains 2.00 mole of CH_4 .3.0 mole of CS_2 ,3.0 mole of H_2 and 4.0 mole of H_2S .Then

    A
    This reaction is at equilibrium with above concentrations
    B
    The reaction will proceed in forward direction to reach equilibrium
    C
    The reaction will proceed in backward direction to reach equilibrium
    D
    The information is insufficient to decide the direction of progress of reaction

  • What is the unit of K_(p) for the reaction ? CS_(2)(g)+4H_(4)(g)hArrCH_(4)(g)+2H_(2)S(g)

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    atm
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    `atm^(-2)`
    C
    `atm^(2)`
    D
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  • What is the unit of K_p for the reaction? CS_2 (g)+4H_2 (g) harr CH_4(g) +2H_2S(g)

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