Question

What is the change in enthalpy for the single replacement reaction between solid zinc and Cs2SO4 in solution to produce cesium and ZnSO4? Zn + Cs2SO4 → 2Cs + ZnSO4 Given: `"Cs"_2"SO"_4(aq): DeltaH_f °= –1,400 "kJ"` `"ZnSO"_4(aq): DeltaH_f °= –1,063 "kJ"`

Answer 1

The change in enthalpy refers to the amount of heat absorbed or heat evolved in a reaction which is carried out at constant pressure.

The change in enthalpy of a reaction is equal to the sum of formation of products minus sum of formation of reactants. It is denoted by $\Delta H$.

The given reaction is:

$Zn + Cs_{2}SO_{4} \rightarrow 2Cs + ZnSO_{4}$

$\Delta H = \Delta H_{f}^{0}(products))-\Delta H_{f}^{0}(reactants))$

Substitute the given values $\Delta H_{f}^{0}$(products)  and $\Delta H_{f}^{0}$(reactants) in above formula, we get

$\Delta H = -1,063 kJ -(-1,400 kJ)$

= $+ 337 kJ$

Thus, change in enthalpy for the single replacement reaction is $+ 337 kJ$