To find the value of the equilibrium constant [tex]\( K_c \)[/tex] for the reaction
[tex]\[ B_2(g) \rightleftharpoons 2B(g), \][/tex]
we need to use the given rate constants [tex]\( k_{\text{fwd}} \)[/tex] and [tex]\( k_{\text{rev}} \)[/tex].
1. The forward rate constant [tex]\( k_{\text{fwd}} \)[/tex] is given as [tex]\( 7.00 \times 10^{-5} \, \text{s}^{-1} \)[/tex].
2. The reverse rate constant [tex]\( k_{\text{rev}} \)[/tex] is given as [tex]\( 2.00 \times 10^{-5} \, \text{L mol}^{-1} \text{s}^{-1} \)[/tex].
The equilibrium constant [tex]\( K_c \)[/tex] is defined as the ratio of the forward rate constant to the reverse rate constant:
[tex]\[ K_c = \frac{k_{\text{fwd}}}{k_{\text{rev}}}. \][/tex]
Substituting the given values, we have:
[tex]\[ K_c = \frac{7.00 \times 10^{-5} \, \text{s}^{-1}}{2.00 \times 10^{-5} \, \text{L mol}^{-1} \text{s}^{-1}}. \][/tex]
Performing the division:
[tex]\[ K_c = \frac{7.00}{2.00} \times \frac{10^{-5}}{10^{-5}}. \][/tex]
Since [tex]\( 10^{-5} \)[/tex] in the numerator and denominator cancel each other out, we are left with:
[tex]\[ K_c = \frac{7.00}{2.00} = 3.50. \][/tex]
Thus, the value of the equilibrium constant [tex]\( K_c \)[/tex] under these conditions is [tex]\( 3.50 \)[/tex].
