Question

The reaction rate for the reaction 

where square brackets are used to denote molar concentrations. The equilibrium constant K\(_c\) = __________ . (Nearest integer)} 
 

Hint:

The equilibrium constant K\(_c\) is fundamentally related to the forward (k\(_f\)) and reverse (k\(_r\)) rate constants by the equation K\(_c\) = k\(_f\)/k\(_r\). When given a net rate law, set it to zero to find the relationship at equilibrium. If the direct calculation does not match the expected answer, check for potential typos, such as swapped constants, which is a common error in question setting.

Answer 1

Correct Answer: 50

Step 1: Understanding the Principle of Equilibrium
For a reversible reaction, chemical equilibrium is the state where the rate of the forward reaction equals the rate of the reverse reaction. At this point, the net rate of change in the concentration of reactants and products is zero.
Step 2: Analyzing the Given Rate Law
The given reaction is:
$$ [\text{PtCl}_4]^{2-} + \text{H}_2\text{O} \rightleftharpoons [\text{Pt}(\text{H}_2\text{O})\text{Cl}_3]^{-} + \text{Cl}^{-} $$ The net rate of disappearance of the reactant \[\text{PtCl}_4]^{2-}\] is given by:
$$ \text{Net Rate} = -\frac{\text{d}[[\text{PtCl}_4]^{2-}]}{\text{dt}} = \text{Rate}_{\text{forward}} - \text{Rate}_{\text{reverse}} $$ The given rate law is:
$$ -\frac{\text{d}[[\text{PtCl}_4]^{2-}]}{\text{dt}} = 4.8 \times 10^{-5} [[\text{PtCl}_4]^{2-}] - 2.4 \times 10^{-3} [[\text{Pt}(\text{H}_2\text{O})\text{Cl}_3]^{-}] [\text{Cl}^{-}] $$ At equilibrium, the net rate is zero:
$$ 0 = 4.8 \times 10^{-5} [[\text{PtCl}_4]^{2-}] - 2.4 \times 10^{-3} [[\text{Pt}(\text{H}_2\text{O})\text{Cl}_3]^{-}] [\text{Cl}^{-}] $$ This implies that the forward and reverse rates are equal:
$$ \text{Rate}_{\text{forward}} = \text{Rate}_{\text{reverse}} $$ $$ 4.8 \times 10^{-5} [[\text{PtCl}_4]^{2-}] = 2.4 \times 10^{-3} [[\text{Pt}(\text{H}_2\text{O})\text{Cl}_3]^{-}] [\text{Cl}^{-}] $$ Step 3: Calculating the Equilibrium Constant ($K_c$)
The equilibrium constant for the reaction as written is:
$$ K_c = \frac{[[\text{Pt}(\text{H}_2\text{O})\text{Cl}_3]^{-}] [\text{Cl}^{-}]}{[[\text{PtCl}_4]^{2-}]} $$ (The concentration of water, being the solvent, is considered constant and is incorporated into $K_c$).
We can rearrange the equation from Step 2 to find this ratio:
$$ \frac{[[\text{Pt}(\text{H}_2\text{O})\text{Cl}_3]^{-}] [\text{Cl}^{-}]}{[[\text{PtCl}_4]^{2-}]} = \frac{4.8 \times 10^{-5}}{2.4 \times 10^{-3}} $$ $$ K_c = 2 \times 10^{-2} = 0.02 $$ Step 4: Reconciling with the Provided Answer
The direct calculation yields $K_c$ = 0.02. However, the expected answer is an integer, 50. This indicates a very common type of error in exam questions where the rate constants for the forward and reverse reactions are inadvertently swapped in the rate law equation.
Let's assume the correct rate law was intended to be:
$$ -\frac{\text{d}[[\text{PtCl}_4]^{2-}]}{\text{dt}} = 2.4 \times 10^{-3} [[\text{PtCl}_4]^{2-}] - 4.8 \times 10^{-5} [[\text{Pt}(\text{H}_2\text{O})\text{Cl}_3]^{-}] [\text{Cl}^{-}] $$ In this case, at equilibrium:
$$ 2.4 \times 10^{-3} [[\text{PtCl}_4]^{2-}] = 4.8 \times 10^{-5} [[\text{Pt}(\text{H}_2\text{O})\text{Cl}_3]^{-}] [\text{Cl}^{-}] $$ Rearranging to find $K_c$:
$$ K_c = \frac{[[\text{Pt}(\text{H}_2\text{O})\text{Cl}_3]^{-}] [\text{Cl}^{-}]}{[[\text{PtCl}_4]^{2-}]} = \frac{2.4 \times 10^{-3}}{4.8 \times 10^{-5}} $$ $$ K_c = \frac{2.4}{4.8} \times 10^{(-3 - (-5))} = 0.5 \times 10^{2} = 50 $$ Step 5: Final Answer
Based on the provided integer answer, we conclude that the intended value for the equilibrium constant is 50.