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Calculate the equilibrium constant (K(c...

Calculate the equilibrium constant `(K_(c))` for the reaction given below , if at equilibrium, mixture contains 5.0 mole of `A_(2)`,3 mole of `B_(2)` and 2 mole of `AB_(2)` at 8.21 atm and 300K
`A_(2)(g)+2B_(2)(g)hArr2AB_(2)(g)+Heat`

A

`1.333`

B

`2.66`

C

`20`

D

none of these

Text Solution

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The correct Answer is:
To calculate the equilibrium constant \( K_c \) for the reaction \[ A_2(g) + 2B_2(g) \rightleftharpoons 2AB_2(g) + \text{Heat} \] given the equilibrium concentrations, we will follow these steps: ### Step 1: Write the expression for the equilibrium constant \( K_c \) The equilibrium constant \( K_c \) for the reaction can be expressed as: \[ K_c = \frac{[AB_2]^2}{[A_2][B_2]^2} \] ### Step 2: Calculate the total number of moles At equilibrium, we have: - Moles of \( A_2 = 5.0 \) moles - Moles of \( B_2 = 3.0 \) moles - Moles of \( AB_2 = 2.0 \) moles Total moles at equilibrium: \[ n_{total} = n_{A_2} + n_{B_2} + n_{AB_2} = 5 + 3 + 2 = 10 \text{ moles} \] ### Step 3: Calculate the volume using the ideal gas law Using the ideal gas law \( PV = nRT \), we can find the volume \( V \): Given: - Total moles \( n = 10 \) - Pressure \( P = 8.21 \) atm - Temperature \( T = 300 \) K - Gas constant \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \) Rearranging the ideal gas law to solve for \( V \): \[ V = \frac{nRT}{P} = \frac{10 \times 0.0821 \times 300}{8.21} \] Calculating \( V \): \[ V = \frac{246.3}{8.21} \approx 30 \, \text{L} \] ### Step 4: Calculate the concentrations Now we can calculate the concentrations of each component: \[ [A_2] = \frac{n_{A_2}}{V} = \frac{5.0}{30} = 0.1667 \, \text{mol/L} \] \[ [B_2] = \frac{n_{B_2}}{V} = \frac{3.0}{30} = 0.1 \, \text{mol/L} \] \[ [AB_2] = \frac{n_{AB_2}}{V} = \frac{2.0}{30} = 0.0667 \, \text{mol/L} \] ### Step 5: Substitute the concentrations into the \( K_c \) expression Now substituting the concentrations into the \( K_c \) expression: \[ K_c = \frac{(0.0667)^2}{(0.1667)(0.1)^2} \] Calculating \( K_c \): \[ K_c = \frac{0.004445}{0.1667 \times 0.01} = \frac{0.004445}{0.001667} \approx 2.67 \] ### Final Answer Thus, the equilibrium constant \( K_c \) for the reaction is approximately: \[ K_c \approx 2.67 \] ---
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