Question

a chemist measures the enthalpy change δh during the following reaction: fe(s) + 2 hcl(g) → fecl₂(s) + h₂(g) δh = -157. kj use this information to complete the table below. round each of your answers to the nearest kj. reaction δh 2fecl₂(s) + 2h₂(g) → 2fe(s) + 4hcl(g) kj 3fe(s) + 6hcl(g) → 3fecl₂(s) + 3h₂(g) kj fecl₂(s) + h₂(g) → fe(s) + 2hcl(g) kj

Explanation:

Step1: Analyze first reaction

Reverse original reaction and multiply by 2: $\Delta H = -(-157) \times 2 = 314$ kJ

Step2: Analyze second reaction

Multiply original reaction by 3: $\Delta H = -157 \times 3 = -471$ kJ

Step3: Analyze third reaction

Reverse original reaction: $\Delta H = -(-157) = 157$ kJ

Answer:

reaction	ΔH
3Fe(s) + 6HCl(g) → 3FeCl₂(s) + 3H₂(g)	-471 kJ
FeCl₂(s) + H₂(g) → Fe(s) + 2HCl(g)	157 kJ