Given : HCl _ (g) + 40 H _2 O HCl (40 H _2 O ); H = -73.0 kJ HCl _ (g) + 10 H _2 O…

CHEMISTRY · Single correct · A Level · Chemistry · Energetics · Energetics

Problem

Given : HCl(g)+40H2O⟶HCl(40H2O);ΔH=−73.0 kJ\text{HCl}_{(g)} + 40\text{H}_2\text{O} \longrightarrow \text{HCl}(40\text{H}_2\text{O}); \Delta H = -73.0 \text{ kJ} HCl(g)+10H2O⟶HCl(10H2O);ΔH=−69.0 kJ\text{HCl}_{(g)} + 10\text{H}_2\text{O} \longrightarrow \text{HCl}(10\text{H}_2\text{O}); \Delta H = -69.0 \text{ kJ} then integral heat of dilution for two moles of HCl(10H2O)\text{HCl}(10\text{H}_2\text{O}) to HCl(40H2O)\text{HCl}(40\text{H}_2\text{O}) is : (1) -4.5 kJ (2) -3.5 kJ (3) -8.0 kJ (4) -2.0 kJ

Related theory

Browse the problem library