The heat of solution of Na_2SO_{4(s)} and Na_ | vedclass

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(A) The heat of solution for $Na_2SO_{4(s)}$ is given by: $Na_2SO_{4(s)} + aq \rightarrow Na_2SO_{4(aq)}$,$\Delta H_1 = -2.34 \ kJ \ mol^{-1} \dots (i

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$-81.21 \ kJ \ mol^{-1}$

Vedclass · Answer

(A) The heat of solution for $Na_2SO_{4(s)}$ is given by: $Na_2SO_{4(s)} + aq \rightarrow Na_2SO_{4(aq)}$,$\Delta H_1 = -2.34 \ kJ \ mol^{-1} \dots (i)$ The heat of solution for $Na_2SO_4 \cdot 10H_2O_{(s)}$ is given by: $Na_2SO_4 \cdot 10H_2O_{(s)} + aq \rightarrow Na_2SO_{4(aq)}$,$\Delta H_2 = 78.87 \ kJ \ mol^{-1} \dots (ii)$ The heat of hydration is the enthalpy change for the reaction: $Na_2SO_{4(s)} + 10H_2O_{(\ell)} \rightarrow Na_2SO_4 \cdot 10H_2O_{(s)}$ This can be obtained by subtracting equation $(ii)$ from equation $(i)$: $\Delta H_{hydration} = \Delta H_1 - \Delta H_2$ $\Delta H_{hydration} = -2.34 \ kJ \ mol^{-1} - 78.87 \ kJ \ mol^{-1} = -81.21 \ kJ \ mol^{-1}$

The heat of solution of $Na_2SO_{4(s)}$ and $Na_2SO_4 \cdot 10H_2O_{(s)}$ are $-2.34 \ kJ \ mol^{-1}$ and $78.87 \ kJ \ mol^{-1}$ respectively. The heat of hydration of $Na_2SO_{4(s)}$ is:

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