Problem

Write the equilibrium expression, $K_{b}$, for the reaction represented above.

Answer

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Answer

\(\boxed{K_b = 0}\)

Steps

Step 1 ：\(K_b = \frac{[B]^2}{[A]^2}\)

Step 2 ：\(K_b = \frac{(b + 5)^2}{(2b - 3)^2}\)

Step 3 ：\(K_b = \frac{(-5 + 5)^2}{(2(-5) - 3)^2}\)

Step 4 ：\(K_b = \frac{0^2}{(-13)^2}\)

Step 5 ：\(K_b = \frac{0}{169}\)

Step 6 ：\(\boxed{K_b = 0}\)

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