SATHEE: Equilibrium Question 623

Equilibrium Question 623

Question: In a chemical equilibrium, the rate constant of the backward reaction is $7.5 \times 10^{- 4}$ and the equilibrium constant is 1.5. So the rate constant of the forward reaction is [KCET 1989]

Options:

A) $5 \times 10^{- 4}$

B) $2 \times 10^{- 3}$

C) $1.125 \times 10^{- 3}$

D) $9.0 \times 10^{- 4}$

Show Answer

Answer:

Correct Answer: C

Solution:

$K_{c} = \frac{K_{f}}{K_{b}}$

$K_{f} = K_{c} \times K_{b} = 1.5 \times 7.5 \times 10^{- 4}$

$= 1.125 \times 10^{- 3}$

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Equilibrium Question 623

Question: In a chemical equilibrium, the rate constant of the backward reaction is 7.5×10−4 and the equilibrium constant is 1.5. So the rate constant of the forward reaction is [KCET 1989]

Options:

Answer:

Solution:

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Question: In a chemical equilibrium, the rate constant of the backward reaction is $7.5 \times 10^{- 4}$ and the equilibrium constant is 1.5. So the rate constant of the forward reaction is [KCET 1989]