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 6270 C- Which of the following options is-are correct if K c -6-30 - 10-6 - Given- -R -0-08 atm - L - molK A- Reaction quotient is 8 - 10-4B- Reaction proceeds in backward direction C- K p -2-35 atm 3D- Reaction proceeds in forward direction

The correct options are A Reaction quotient is nbsp;8× 10^ 4 B Reaction proceeds in backward direction C K_p=2.35 atm^3Reaction quotient Q=[S_2]^4/[S_8]=(0.2) ...

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Byju's Answer

Standard XII

Chemistry

Heat Capacity at Constant Pressure

A reaction S ...

Question

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 $a t m - L / m o l K$ )

Reaction quotient is

8 \times 10^{- 4}

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Reaction proceeds in backward direction

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K_{p} = 2.35 a t m^{3}

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Reaction proceeds in forward direction

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Solution

The correct options are
A Reaction quotient is $8 \times 10^{- 4}$
B Reaction proceeds in backward direction
C $K_{p} = 2.35 a t m^{3}$
Reaction quotient $Q = \frac{[S_{2}]^{4}}{[S_{8}]} = \frac{(0.2)^{4}}{2} = 8 \times 10^{- 4}$
As $Q > K_{c}$ , reaction will go backwards.
$K_{p} = K_{c} \times (R T)^{Δ n_{g}}$
$K_{p} = 6.3 \times 10^{- 6} \times (0.08 \times 900)^{3} = 2.35 a t m^{3}$

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A reaction S8(g)⇌4S2(g) is carried out by taking 2 moles of S8(g) and 0.2 mole of of S2 in the reaction vessel of 1 litre and temperature 627 0C. Which of the following option(s) is/are correct if Kc= 6.30×10−6? (Given: R= 0.08 atm−L/molK)

The correct options are A Reaction quotient is 8×10−4 B Reaction proceeds in backward direction C Kp=2.35 atm3Reaction quotient Q=[S2]4[S8]=(0.2)42=8×10−4 As Q>Kc, reaction will go backwards. Kp=Kc×(RT)Δng Kp=6.3×10−6×(0.08×900)3=2.35 atm3