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Multiple Choice
Consider the reaction A(aq) ⇌ B(aq) in an equilibrium state where K_C < 1. If the rate of the forward reaction A(aq) ⇌ B(aq) at equilibrium is 59.9 mol/s, what is the rate of the reverse reaction B(aq) ⇌ A(aq) at equilibrium?
A
0 mol/s
B
Less than 59.9 mol/s
C
Greater than 59.9 mol/s
D
59.9 mol/s
Verified step by step guidance
1
Understand that at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. This is a fundamental concept in chemical equilibrium.
Identify that the equilibrium constant \( K_C \) is less than 1, which indicates that the concentration of reactants \( A(aq) \) is greater than the concentration of products \( B(aq) \) at equilibrium.
Recognize that the rate of the forward reaction \( A(aq) \rightarrow B(aq) \) is given as 59.9 mol/s at equilibrium.
Since the system is at equilibrium, the rate of the reverse reaction \( B(aq) \rightarrow A(aq) \) must also be 59.9 mol/s to maintain the equilibrium state.
Conclude that the rate of the reverse reaction is equal to the rate of the forward reaction at equilibrium, which is 59.9 mol/s.