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Standard XII

Chemistry

Reaction Quotient

A reaction ...

Question

A reaction $S_{8} (g) ⇌ 4 S_{2} (g)$ is carried out by taking 2 mol of $S_{8} (g)$ and 0.2 mol of $S_{2} (g)$ is a reaction vessel of 1 L. Which one is not correct if $K_{c} = 6.30 \times 10^{- 6}$ ?

Reaction qutient is

8 \times 10^{- 4}

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Reaction proceed in backward direction

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Reaction proceed in forward direction

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K_{p} = 2.55 a t m^{3}

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Solution

The correct options are
A Reaction proceed in backward direction
C $K_{p} = 2.55 a t m^{3}$
D Reaction qutient is $8 \times 10^{- 4}$
$Q = \frac{[S_{2}]^{4}}{[S_{8}]} = \frac{(0.2)^{4}}{(2)}$
$= 8 \times 10^{- 4};$
$K_{c} = 6.3 \times 10^{- 6}$
$∴ K_{p} = K_{c} \times (R T)^{Δ n}$
$= 6.3 \times 0^{- 6} \times (0.0821 \times 900)^{3}$
$= 2.55$
Thus, reaction proceed in backward direction since $Q > K_{c}$ .

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Similar questions

Q. A reaction

S_{8} (g) ⇌ 4 S_{2} (g)

is carried out by taking 2 moles of

S_{8} (g)

and 0.2 mole of of

S_{2}

in the reaction vessel of 1 litre and temperature

627^{0} C

. Which of the following option(s) is/are correct if

K_{c} = 6.30 \times 10^{- 6}

? (Given: R= 0.08

a t m - L / m o l K

)

Q. When sulphur in the form of

S_{2} (g)

is heated at

900 K

, the initial partial pressure of

S_{8} (g)

which was

1 a t m

falls by

29 %

at equilibrium. This is because of conversion of some

S_{8} (g)

S_{2} (g)

. Find the

K_{p}

for reaction,

S_{8} (g) ⇌ 4 S_{2} (g)

Q. The value of

K_{c}

for the reaction

2 H I (g) ⇌ H_{2} (g) + I_{2} (g)

1 \times 10^{- 4}

. At a given time, the composition of reaction mixture is

[H I] = 2 \times 10^{- 5}

mol and

[I_{2}] = 1 \times 10^{- 5}

mol. The direction in which the reaction will proceed is (For forward direction write 0 and for reverse direction write 1)

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. 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:

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A reaction S8(g)⇌4S2(g) is carried out by taking 2 mol of S8(g) and 0.2 mol of S2(g) is a reaction vessel of 1 L. Which one is not correct if Kc=6.30×10−6?

The correct options are A Reaction proceed in backward direction C Kp=2.55atm3 D Reaction qutient is 8×10−4Q=[S2]4[S8]=(0.2)4(2)=8×10−4;Kc=6.3×10−6∴Kp=Kc×(RT)Δn=6.3×0−6×(0.0821×900)3=2.55Thus, reaction proceed in backward direction since Q>Kc.

A reaction $S_{8} (g) ⇌ 4 S_{2} (g)$ is carried out by taking 2 mol of $S_{8} (g)$ and 0.2 mol of $S_{2} (g)$ is a reaction vessel of 1 L. Which one is not correct if $K_{c} = 6.30 \times 10^{- 6}$ ?