Consider the reaction
$A (g) + B (g) ⇌ C (g) + D (g)$
Which occurs in one step The specific rate constant are $0.25$ and $5000$ for the forward and reverse reaction, respectively. The equilibrium constant is :

2.0 \times 10^{- 4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4.0 \times 10^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5.0 \times 10^{- 5}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2.5 \times 10^{- 6}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $5.0 \times 10^{- 5}$
As we know,
$K = \frac{R_{f}}{R_{b}} = \frac{0.25}{5000} = 5.0 \times 10^{- 5}$

Suggest Corrections

Similar questions

Consider the reaction.

$A_{(g)} + B_{(g)} ⇌ C_{(g)} + D_{(g)}$ which occurs in one step. The specific rate constants are 0.25 and 5000 for the forward and reverse reaction, respectively. The equilibrium constant is

Q. The equilibrium constant

K_{c}

for,

A (g) ⇋ B (g)

2.5 \times 10^{- 2}

. The rate constant of forward reaction is

0.05 s e c^{- 1}

. Therefore, the rate constant of the reverse reaction is:

A hypothetical reaction : $A_{(g)} + B_{(g)} ⇌ C_{(g)} + D_{(g)}$ occurs in a single step, the specific rate constant of forward reaction at TK is $2.0 \times 10^{- 3} m o l^{- 1} L s^{- 1}$ . When start is made with equimolar amounts of A and B, it is found that the concentration of A is twice that of C at equilibrium. The specific rate constant of the backward reaction is ?

Q. For the reaction,

A_{(g)} + 2 B_{(g)} ⇌ 2 C_{(g)}

, the rate constants for the forward and the reverse reactions are

1 \times 10^{- 4}

and

2.5 \times 10^{- 2}

respectively. The value of equilibrium constant, K for the reaction would be-

Join BYJU'S Learning Program

Consider the reactionA(g)+B(g)⇌C(g)+D(g)Which occurs in one step The specific rate constant are 0.25 and 5000 for the forward and reverse reaction, respectively. The equilibrium constant is :

The correct option is C 5.0×10−5As we know,K=RfRb=0.255000=5.0×10−5

Consider the reaction
$A (g) + B (g) ⇌ C (g) + D (g)$
Which occurs in one step The specific rate constant are $0.25$ and $5000$ for the forward and reverse reaction, respectively. The equilibrium constant is :

The correct option is C $5.0 \times 10^{- 5}$
As we know,
$K = \frac{R_{f}}{R_{b}} = \frac{0.25}{5000} = 5.0 \times 10^{- 5}$