# ICE table with changing volume without Kc [closed]

Just having trouble understanding how to approach an ICE table question with no Kc and a volume change

Given

If you had 4.4 mol of a solid, A, placed in a 1L container which decomposed in to B and C (gas), B would slowly increase until 1.20M

reaction: $$\ce{A(s) <=> B(g) + C(g)}$$

Question

If the volume doubled and equilibrium was re-established, how many mol of A would there be?

My attempt:

Kc = 1.2M + 1.2M (not including solids)

Kc = 2.4M

for new ice table

2.4 = ((1.2 + x)(1.2 + x)) / 2 (because now litres is doubled)

0 = x^2 + 2.4x - 3.36

x = 0.9908 (which is wrong apparently)

• Welcome to StackExchange Chemistry. Do you know what the equilibrium constant expression is for this reaction. You could use the first equilibrium to calculate Kc and then apply it to the second equilibrium. Could you edit your question to include how far you got, and where specifically you got stuck? – Karsten Theis Apr 23 at 3:25
• David, were you given the reaction or did you propose it? – MaxW Apr 23 at 3:36
• This reaction was given to me – David Apr 23 at 3:37
• How you calculated the Kc is wrong. The Kc is the equilibrium constant using concentrations. // Frankly using an ICE table is overkill and confusing as far as I am concerned. – MaxW Apr 23 at 3:42
• Why are you adding values to get $K_{\mathrm{c}}$? I'm hard pressed to come up with even a contrived situation where that would be the correct approach. – Zhe Apr 23 at 13:18

For the numerical result, see the answer by MaxW.

Kc = 1.2M + 1.2M (not including solids)

The concentrations of B and C should be multiplied, not added. So Kc is 1.2 squared.

2.4 = ((1.2 + x)(1.2 + x)) / 2 (because now litres is doubled)

$$K_c = \frac{1.2 + x}{2} \times \frac{1.2 + x}{2}$$

Why you don't need an ICE table

This problem is similar to an ionic solid dissolving:

$$\ce{NaCl(s) -> Na+(aq) + Cl-(aq)}$$

This reaches equilibrium when the solution is saturated. If you add more water (dilute it), more will dissolve until the solution is saturated again or you run out of solid. If you don't run out of solid, the concentration before dilution and after dilution will be the same (that concentration, based on the amount of NaCl dissolved, is called solubility, and the equilibrium constant for the process is called solubility product).

Let’s look at this at different approach. Suppose your initial volume is $$V$$. $$\begin{array}{ccc} \text{Reaction:} & \ce{A (s) & <=> & B (g) & + & C (g)}\\ \hline \text{initial amount:} & \pu{4.4 mol} & & \pu{0 mol} & & \pu{0 mol} \\ \text{amount at equilibrium:} & \text{some solid} & & \pu{1.20 mol} & & \pu{1.20 mol} \\ \text{equilibrium concentration:} & \text{some solid} & & \pu{\frac{1.20}{V} mol\: L^{-1}} & & \pu{\frac{1.20}{V} mol\: L^{-1}} \\ \text{amount at equilibrium at 2V:} & \text{some solid} & & \pu{x mol} & & \pu{x mol} \\ \text{equilibrium concentration at 2V:} & \text{some solid} & & \pu{\frac{x}{2V} mol\: L^{-1}} & & \pu{\frac{x}{2V} mol\: L^{-1}} \\ \hline \end{array}$$ Thus, $$K_\mathrm{C}$$ when volume is $$V$$ can be derived as:
$$K_\mathrm{C} = \pu{\frac{1.20}{V} mol\: L^{-1}} \times \pu{\frac{1.20}{V} mol\: L^{-1}} = \pu{\left(\frac{1.20}{V}\right)^2 mol^2 L^{-2}}$$
Similarly, because it is the same reaction, $$K_\mathrm{C}$$ can also be derived when volume is $$2V$$ as:
$$K_\mathrm{C} = \pu{\frac{x}{2V} mol\: L^{-1}} \times \pu{\frac{x}{2V} mol\: L^{-1}} = \pu{\left(\frac{x}{2V}\right)^2 mol^2 L^{-2}}$$ As a result, $$K_\mathrm{C} = \pu{\left(\frac{1.20}{V}\right)^2 mol^2 L^{-2}} = \pu{\left(\frac{x}{2V}\right)^2 mol^2 L^{-2}}$$ Solving this equation, you may calculate values for $$x$$: $$x = \pu{1.20 \times 2 mol}= \pu{2.40 mol}$$