If the equilibrium constant for a reverse reaction is 9-0- 10 -4 is the equilibrium constant for the forward reaction-

The correct option is D 1/9.0× 10 ^ 4 #160; K_eq(forward)×K_eq(reverse) =1 K_eq(forward)× #160; 9.0 × 10^ 4= 1 K_eq(forward) = 1 9.0 × ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Characteristics of Equilibrium Constant

If the equili...

Question

If the equilibrium constant for a reverse reaction is $9.0 \times 10^{- 4}$ is the equilibrium constant for the forward reaction?

3.0 \times 10^{- 2}

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

- 3.0 \times 10^{- 2}

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

- 9.0 \times 10^{- 2}

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

1 / 9.0 \times 10^{- 4}

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

1 / - 9.0 \times 10^{- 4}

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

Open in App

Solution

The correct option is D $1 / 9.0 \times 10^{- 4}$
$K_{e q} (f o r w a r d) \times K_{e q} (r e v e r s e) = 1$
$K_{e q} (f o r w a r d) \times 9.0 \times 10^{- 4} = 1$
$K_{e q} (f o r w a r d) = \frac{1}{9.0 \times 10^{- 4}}$

Suggest Corrections

Similar questions

A coil of wire has an area of 2.0x10^-4m², consists of 100 loops or turns, and contains a current of 0.045 A. The coil is placed in a uniform magnetic field of magnitude 0.15 T. Determine the magnetic moment of the coil.

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-

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. 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. The dissociation constants for acetic acid and HCN at

25^{\circ} C

are

1.5 \times 10^{- 5}

and

4.5 \times 10^{- 10}

, respectively. The equilibrium constant for the equilibrium,

C N^{-} + C H_{3} C O O H ⇌ H C N + C H_{3} C O O^{-}

would be?

Join BYJU'S Learning Program

If the equilibrium constant for a reverse reaction is 9.0×10−4 is the equilibrium constant for the forward reaction?

The correct option is D 1/9.0×10−4Keq(forward)×Keq(reverse)=1Keq(forward)×9.0×10−4=1Keq(forward)=19.0×10−4

If the equilibrium constant for a reverse reaction is $9.0 \times 10^{- 4}$ is the equilibrium constant for the forward reaction?

The correct option is D $1 / 9.0 \times 10^{- 4}$
$K_{e q} (f o r w a r d) \times K_{e q} (r e v e r s e) = 1$
$K_{e q} (f o r w a r d) \times 9.0 \times 10^{- 4} = 1$
$K_{e q} (f o r w a r d) = \frac{1}{9.0 \times 10^{- 4}}$