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

337 kJ

Step-by-step explanation:

Plato

Answer 2

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`