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Svet_ta [14]
3 years ago
8

At 700 K the equilibrium constant KC for the reaction between NO(g) and O2(g) forming NO2(g) is 8.7 × 106. The rate constant for

the reverse reaction at this temperature is 0.54 M–1s–1. What is the value of the rate constant for the forward reaction at 700 K?
Chemistry
1 answer:
VMariaS [17]3 years ago
7 0

Answer : The value of the rate constant for the forward reaction at 700 K is, 4.70\times 10^6

Explanation :

The given chemical equilibrium reaction is:

NO(g)+O_2(g)\rightleftharpoons NO_2(g)

The expression for equilibrium constant is:

K_c=\frac{[NO_2]}{[NO][O_2]}

The expression for rate of forward and backward reaction is:

R_f=K_f[NO][O_2]

and,

R_b=K_b[NO_2]

As we know that at equilibrium rate of forward reaction is equal to rate of backward reaction.

R_f=R_b

K_f[NO][O_2]=K_b[NO_2]

\frac{K_f}{K_b}=\frac{[NO_2]}{[NO][O_2]}

\frac{K_f}{K_b}=K_c

Given:

K_c=8.7\times 10^6

K_b=0.54M^{-1}s^{-1}

Now put all the given values in the above expression we get:

\frac{K_f}{K_b}=K_c

\frac{K_f}{0.54}=8.7\times 10^6

K_f=4.70\times 10^6

Therefore, the value of the rate constant for the forward reaction at 700 K is, 4.70\times 10^6

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