Question

the rate of constant forward reaction in a reversible reaction is 10^2​

Answer 1

Answer:

We have, K

eq

=

K

b

K

f

Now, K

f

=2.38×10

−4

;K

b

=8.15×10

−5

⇒K

eq

=

8.15×10

−5

2.38×10

−4

=2.92

Answer 2

Answer:

The rate constant for forward reaction A(g)⇔2B(g) is 1.5×10-3s-1 at 100K. If 10-5 moles of A and 100moles of B are present in a 10-L vessel at equilibrium, then the rate constant for the backward reaction at this temperature is.