Question

if the equilibrium constant for reaction 1/2A2+1/2B2⇌AB is k then equilibrium constant for the reaction A2+B2⇌2AB is​

Answer 1

Explanation:

pH=6

∴H

+

=10

−6

M

After dilution [H

+

]=

1000

10

−6

=10

−9

M

∴[H

+

] from H

2

O cannot be neglected.

Total [H

+

]=10

−9

+10

−7

=10

−7

(10

−2

+1)

=10

−7

(1.01)

pH=−log(1.01×10

−7

)

=7−0.0043

=6.9957

Answer 2

Given: The equilibrium constant for reaction 1/2A2+1/2B2⇌AB is k.

To find: We have to find the equilibrium constant for $A_2+B_2⇌2AB$.

Solution:

The given reaction is $1/2A_2+1/2B_2⇌AB$.

The equilibrium constant for this reaction is k.

k is given as-

$k=\dfrac {[AB]}{[A_2]^{1/2}[B_2]^{1/2}}$.

The other reaction is $A_2+B_2⇌2AB$.

Let the equilibrium constant for this reaction be K.

So, the equilibrium constant is given as-

$K=\dfrac {[AB]^2}{[A_2][B_2]}$

Taking square root in both sides we get-

$√K=\dfrac {[AB]}{[A_2]^{1/2}[B_2]^{1/2}}$

$√K=k$

$K=k^2$

The value of the equilibrium constant is $k^2$.