Question

the standard enthalpy change for the following reaction is -265 kj at 298 k. 2 hg(l) + cl2(g)→hg2cl2(s) δh° = -265 kj what is the standard enthalpy change for the reaction hg2cl2(s)→2 hg(l) + cl2(g) at 298 k? use the references to access important values if needed for this question.

Explanation:

Step1: Reverse the reaction

The given reaction is \(2Hg(l)+Cl_2(g)
ightarrow Hg_2Cl_2(s)\) with \(\Delta H^{\circ}=- 265\ kJ\). We want \(Hg_2Cl_2(s)
ightarrow2Hg(l)+Cl_2(g)\). When we reverse a reaction, the sign of the enthalpy - change changes.

Step2: Determine the new \(\Delta H^{\circ}\)

If \(\Delta H^{\circ}\) for \(2Hg(l)+Cl_2(g)
ightarrow Hg_2Cl_2(s)\) is \(-265\ kJ\), then for \(Hg_2Cl_2(s)
ightarrow2Hg(l)+Cl_2(g)\), \(\Delta H^{\circ}=+265\ kJ\).

Answer:

\(265\)