Reversible Reactions and Equilibrium: Question 1

Worked solution

Step 1: Identify what the question is really testing

The question describes a change that goes both ways: heating drives off water (pink $\rightarrow$ blue), and adding water reverses it (blue $\rightarrow$ pink). An equation for this change must show that it is reversible, not just balanced.

Step 2: Check each option against this requirement

• Option A: $\text{CoCl}_2 \cdot 6\text{H}_2\text{O(s)} \rightarrow \text{CoCl}_2\text{(s)} + 6\text{H}_2\text{O(g)}$ has the correct formulae and is balanced, but the single arrow $\rightarrow$ only shows the dehydration (heating) direction. It does not show that the change can go backwards.
• Option B: $\text{CoCl}_2 \cdot 6\text{H}_2\text{O(s)} \rightleftharpoons \text{CoCl}_2\text{(s)} + 6\text{H}_2\text{O(g)}$ has the same correct, balanced formulae and uses $\rightleftharpoons$, correctly showing that the reaction can proceed in either direction.
• Option C: $\text{CoCl}_2\text{(s)} + 6\text{H}_2\text{O(g)} \rightarrow \text{CoCl}_2 \cdot 6\text{H}_2\text{O(s)}$ is balanced, but again uses a one-way arrow, this time showing only the rehydration direction.
• Option D: $\text{CoCl}_2 \cdot 6\text{H}_2\text{O(s)} \rightleftharpoons \text{Co(s)} + \text{Cl}_2\text{(g)} + 6\text{H}_2\text{O(g)}$ uses $\rightleftharpoons$, but the products are wrong: heating hydrated cobalt(II) chloride only drives off water of crystallisation, it does not break the compound down into cobalt metal and chlorine gas.

Step 3: Select the correct equation

Only option B has both the correct formulae/balancing and the $\rightleftharpoons$ symbol needed to represent a reversible reaction.

Final answer

• The correctly written reversible equation is $\boxed{\text{CoCl}_2 \cdot 6\text{H}_2\text{O(s)} \rightleftharpoons \text{CoCl}_2\text{(s)} + 6\text{H}_2\text{O(g)}}$, option B.