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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"`

2 Answers

4 votes

Answer:

337 kJ

Step-by-step explanation:

Plato

User Sylence
by
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3 votes

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


User Peter Stuifzand
by
5.4k points