For the reaction- Ag-2Bg- 2Cg- the rate constants for the forward and the reverse reactions are 1- 10-4 and 2-5- 10-2 respectively- The value of equilibrium constant- K for the reaction would be-

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Effective Equilibrium Constant

For the react...

Question

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-

1 \times 10^{- 4}

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

2.5 \times 10^{- 2}

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

4 \times 10^{- 3}

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

2.5 \times 10^{2}

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

Open in App

Solution

The correct option is C $4 \times 10^{- 3}$
The equilibrium reaction is $A_{(g)} + 2 B_{(g)} ⇌ 2 C_{(g)} .$
The expression for the equilibrium constant is $K = \frac{k_{f}}{k_{b}} .$
Substitute values in the above expression.
$K = \frac{1 \times 10^{- 4}}{2.5 \times 10^{- 2}} = 4 \times 10^{- 3} .$

Suggest Corrections

Similar questions

Q. For the reaction,

A + B ⇌ C

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

4 \times 10^{2}

and

2 \times 10^{2}

respectively.

The value of equilibrium constant K for the reaction would be :

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:

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. For a reversible reaction, the rate constants for the forward and backward reactions are

2.38 \times 10^{- 4}

and

8.15 \times 10^{- 5}

respectively. The equilibrium constant for the reaction is:

Q. For an equilibrium reaction, the rate constants for the forward and backward reactions are

2.38 \times 10^{- 4} s^{- 1}

and

8.15 \times 10^{- 5} s^{- 1}

respectively. The equilibrium constant for the reaction is:

Join BYJU'S Learning Program

For the reaction, A(g)+2B(g)⇌2C(g), the rate constants for the forward and the reverse reactions are 1×10−4 and 2.5×10−2 respectively. The value of equilibrium constant, K for the reaction would be-

The correct option is C 4×10−3The equilibrium reaction is A(g)+2B(g)⇌2C(g).The expression for the equilibrium constant is K=kfkb.Substitute values in the above expression.K=1×10−42.5×10−2=4×10−3.

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-