Questions: Part A A mixture initially contains A, B, and C in the following concentrations: [A]=0.650 M, [B]=0.600 M, and [C]= 0.300 M. The following reaction occurs and equilibrium is established: A + 2 B ↔ C At equilibrium, [A]=0.490 M and [C]=0.460 M. Calculate the value of the equilibrium constant, Kc. Express your answer numerically.

Part A

A mixture initially contains A, B, and C in the following concentrations: [A]=0.650 M, [B]=0.600 M, and [C]= 0.300 M. The following reaction occurs and equilibrium is established:
A + 2 B ↔ C

At equilibrium, [A]=0.490 M and [C]=0.460 M. Calculate the value of the equilibrium constant, Kc.
Express your answer numerically.

Transcript text: Part A A mixture initially contains $\mathrm{A}, \mathrm{B}$, and C in the foll concentrations: $[\mathrm{A}]=0.650 \mathrm{M},[\mathrm{B}]=0.600 \mathrm{M}$, and $[\mathrm{C}]=$ 0.300 M . The following reaction occurs and equilibrium is established: \[ \mathrm{A}+2 \mathrm{~B} \rightleftharpoons \mathrm{C} \] At equilibrium, $[\mathrm{A}]=0.490 \mathrm{M}$ and $[\mathrm{C}]=0.460 \mathrm{M}$. Calculate the value of the equilibrium constant, $K_{\mathrm{c}}$. Express your answer numerically. View Available Hint(s) \[ K_{\mathrm{c}}= \] Submit Part B Complete previous part(s) Settings Provide Feedback Next >

Solution

Solution Steps

Step 1: Write the balanced chemical equation and initial concentrations

The balanced chemical equation is: \[ \mathrm{A} + 2\mathrm{B} \rightleftharpoons \mathrm{C} \] The initial concentrations are: \[ [\mathrm{A}]_0 = 0.650 \, \mathrm{M}, \quad [\mathrm{B}]_0 = 0.600 \, \mathrm{M}, \quad [\mathrm{C}]_0 = 0.300 \, \mathrm{M} \]

Step 2: Determine the change in concentration

At equilibrium, the concentration of $\mathrm{A}$ is given as $0.490 \, \mathrm{M}$. The change in concentration of $\mathrm{A}$ is: \[ \Delta [\mathrm{A}] = [\mathrm{A}]_0 - [\mathrm{A}]_{\text{eq}} = 0.650 \, \mathrm{M} - 0.490 \, \mathrm{M} = 0.160 \, \mathrm{M} \] Since the stoichiometry of the reaction shows that 1 mole of $\mathrm{A}$ reacts with 2 moles of $\mathrm{B}$ to form 1 mole of $\mathrm{C}$, the changes in concentrations are: \[ \Delta [\mathrm{B}] = 2 \times \Delta [\mathrm{A}] = 2 \times 0.160 \, \mathrm{M} = 0.320 \, \mathrm{M} \] \[ \Delta [\mathrm{C}] = \Delta [\mathrm{A}] = 0.160 \, \mathrm{M} \]

Step 3: Calculate equilibrium concentrations

Using the changes in concentration, we can find the equilibrium concentrations of $\mathrm{B}$ and $\mathrm{C}$: \[ [\mathrm{B}]_{\text{eq}} = [\mathrm{B}]_0 - \Delta [\mathrm{B}] = 0.600 \, \mathrm{M} - 0.320 \, \mathrm{M} = 0.280 \, \mathrm{M} \] \[ [\mathrm{C}]_{\text{eq}} = [\mathrm{C}]_0 + \Delta [\mathrm{C}] = 0.300 \, \mathrm{M} + 0.160 \, \mathrm{M} = 0.460 \, \mathrm{M} \]

Step 4: Write the expression for the equilibrium constant $K_{\mathrm{c}}$

The equilibrium constant expression for the reaction is: \[ K_{\mathrm{c}} = \frac{[\mathrm{C}]_{\text{eq}}}{[\mathrm{A}]_{\text{eq}} [\mathrm{B}]_{\text{eq}}^2} \]

Step 5: Substitute the equilibrium concentrations into the expression

Substitute the equilibrium concentrations into the expression for $K_{\mathrm{c}}$: \[ K_{\mathrm{c}} = \frac{0.460 \, \mathrm{M}}{(0.490 \, \mathrm{M})(0.280 \, \mathrm{M})^2} \]

Step 6: Calculate the value of $K_{\mathrm{c}}$

Perform the calculation: \[ K_{\mathrm{c}} = \frac{0.460}{(0.490)(0.280)^2} = \frac{0.460}{(0.490)(0.0784)} = \frac{0.460}{0.038416} \approx 11.97 \]