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"`

Respuesta :

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 [tex]\Delta H[/tex].

The given reaction is:

[tex]Zn + Cs_{2}SO_{4} \rightarrow 2Cs + ZnSO_{4}[/tex]

[tex]\Delta H = \Delta H_{f}^{0}(products))-\Delta H_{f}^{0}(reactants))[/tex]

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

[tex]\Delta H = -1,063 kJ -(-1,400 kJ)[/tex]

= [tex]+ 337 kJ[/tex]

Thus, change in enthalpy for the single replacement reaction is [tex]+ 337 kJ[/tex]


Answer:

337 kJ

Explanation:

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