Question

For the reaction A(g) + 2B(g) $\rightleftharpoons$ C(g) + D(g) ; K$_c$ = 10$^{12}$.

If the initial moles of A,B,C and D are 0.5, 1, 0.5 and 3.5 moles respectively in a one litre vessel. What is the equilibrium concentration of B (in millimole/litre) ?

Answer 1

0.2

Answer 2

The reaction is A(g) + 2B(g) $\rightleftharpoons$ C(g) + D(g). The equilibrium constant K$_c$ = 10$^{12}$. The initial moles are: A = 0.5 mol, B = 1 mol, C = 0.5 mol, D = 3.5 mol. The volume of the vessel is 1 L. Therefore, initial concentrations are equal to initial moles.

1. Calculate the reaction quotient (Q$_c$): Q$_c = \frac{[C]_0[D]_0}{[A]_0[B]_0^2} = \frac{(0.5)(3.5)}{(0.5)(1)^2} = \frac{1.75}{0.5} = 3.5$

2. Determine the direction of the reaction: Since Q$_c$ (3.5) < K$_c$ (10$^{12}$), the reaction will proceed in the forward direction to reach equilibrium.

3. Analyze the extent of reaction due to large K$_c$: A very large K$_c$ value (10$^{12}$) indicates that the reaction proceeds almost to completion in the forward direction. Let's determine the limiting reactant for the forward reaction: Initial moles of A = 0.5 mol Initial moles of B = 1 mol According to the stoichiometry (1 mole of A reacts with 2 moles of B): If 0.5 mol of A reacts, it consumes 0.5 * 2 = 1 mol of B. Since we have exactly 0.5 mol of A and 1 mol of B, both reactants A and B will be almost entirely consumed if the reaction goes to completion.

4. Set up an ICE table by assuming complete reaction first, then a small back reaction: Assume the reaction goes to completion in the forward direction: A(g) + 2B(g) $\rightleftharpoons$ C(g) + D(g) Initial (mol): 0.5 1 0.5 3.5 Change (mol): -0.5 -1 +0.5 +0.5 After complete reaction (mol): 0 0 1 4

   Now, since equilibrium must be established (meaning all species are present, even if in very small amounts), a very small amount of products will react backward. Let 'y' be the moles of C that react in the reverse direction to establish equilibrium. C(g) + D(g) $\rightleftharpoons$ A(g) + 2B(g) Initial (after forward completion, mol): 1 4 0 0 Change (mol): -y -y +y +2y Equilibrium (mol): (1-y) (4-y) y 2y

   Since the volume of the vessel is 1 L, these moles are also the equilibrium concentrations in mol/L.

5. Substitute equilibrium concentrations into the K$_c$ expression: K$_c = \frac{[C][D]}{[A][B]^2}$ $10^{12} = \frac{(1-y)(4-y)}{(y)(2y)^2}$ $10^{12} = \frac{(1-y)(4-y)}{4y^3}$

6. Solve for 'y' using approximation: Since K$_c$ is very large, 'y' is expected to be very small. Therefore, we can approximate: (1-y) $\approx$ 1 (4-y) $\approx$ 4 Substituting these approximations: $10^{12} \approx \frac{(1)(4)}{4y^3}$ $10^{12} \approx \frac{4}{4y^3}$ $10^{12} \approx \frac{1}{y^3}$ $y^3 \approx \frac{1}{10^{12}} = 10^{-12}$ $y \approx (10^{-12})^{1/3}$ $y \approx 10^{-4}$ mol/L

7. Calculate the equilibrium concentration of B: From the equilibrium concentrations in the ICE table, [B]$_{eq}$ = 2y. [B]$_{eq}$ = 2 * 10$^{-4}$ mol/L

8. Convert the concentration to millimole/litre: 1 mole = 1000 millimoles [B]$_{eq}$ = 2 * 10$^{-4}$ mol/L * (1000 mmol/mol) [B]$_{eq}$ = 2 * 10$^{-4}$ * 10$^3$ mmol/L [B]$_{eq}$ = 2 * 10$^{-1}$ mmol/L [B]$_{eq}$ = 0.2 mmol/L

The final answer is 0.2.

Explanation of the solution:

1. Calculate initial reaction quotient Q$_c$.
2. Compare Q$_c$ with K$_c$ to determine reaction direction. Q$_c$ < K$_c$, so reaction proceeds forward.
3. Since K$_c$ is very large, the reaction goes almost to completion. Determine limiting reactants (A and B are both limiting).
4. Assume reaction goes to completion, then a small amount 'y' of products react backward to establish equilibrium.
5. Set up equilibrium concentrations in terms of 'y'.
6. Substitute into K$_c$ expression and use approximation (1-y $\approx$ 1, 4-y $\approx$ 4) due to small 'y'.
7. Solve for 'y'.
8. Calculate [B] = 2y.
9. Convert mol/L to millimole/L.