Question

use the references to access important values if needed for this question. potassium sulfate dissolves in water according to the reaction: k₂so₄(s)→2k⁺(aq)+so₄²⁻(aq) a. calculate the standard enthalpy change δh° for this reaction, using the following data: δhf°(k₂so₄(s)) = - 1437.8 kj/mol, δhf°(k⁺(aq)) = - 252.4 kj/mol, δhf°(so₄²⁻(aq)) = - 909.3 kj/mol, δh° = kj b. calculate the temperature reached by the solution formed when 40.2 g of k₂so₄ is dissolved in 0.108 l of water at 21.5 °c. approximate the heat capacity of the solution by the heat capacity of 108 g of pure water (specific heat capacity = 4.18 j/(g·°c)), ignoring the mass of the salt. temperature = °c c. heats of reaction find practical application in hot packs or cold packs. would this dissolution reaction be appropriate for the preparation of a hot pack or a cold pack? hot pack cold pack

Explanation:

Step1: Recall the formula for enthalpy change of a reaction

The formula for the standard - enthalpy change of a reaction $\Delta H^{\circ}=\sum n\Delta H_{f}^{\circ}(products)-\sum m\Delta H_{f}^{\circ}(reactants)$. For the reaction $K_{2}SO_{4}(s)
ightarrow 2K^{+}(aq)+SO_{4}^{2 - }(aq)$, we have $\Delta H^{\circ}=2\Delta H_{f}^{\circ}(K^{+}(aq))+\Delta H_{f}^{\circ}(SO_{4}^{2 - }(aq))-\Delta H_{f}^{\circ}(K_{2}SO_{4}(s))$.

Step2: Substitute the given values

Given $\Delta H_{f}^{\circ}(K^{+}(aq))=- 252.4\ kJ/mol$, $\Delta H_{f}^{\circ}(SO_{4}^{2 - }(aq))=-909.3\ kJ/mol$, and $\Delta H_{f}^{\circ}(K_{2}SO_{4}(s))=-1437.8\ kJ/mol$.

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Step3: Determine if it's a hot - pack or cold - pack reaction

Since $\Delta H^{\circ}=23.7\ kJ/mol>0$, the reaction is endothermic. Endothermic reactions are used in cold - packs as they absorb heat from the surroundings.

Answer:

a. $\Delta H^{\circ}=23.7\ kJ/mol$
b. To find the temperature change, we first need to find the heat absorbed by the solution. The number of moles of $K_{2}SO_{4}$, $n=\frac{m}{M}$, where $m = 40.2\ g$ and $M(K_{2}SO_{4})=(2\times39.1 + 32.07+4\times16.00)=174.27\ g/mol$. So $n=\frac{40.2\ g}{174.27\ g/mol}\approx0.231\ mol$. The heat absorbed by the solution $q = n\Delta H^{\circ}=0.231\ mol\times23.7\ kJ/mol = 5.47\ kJ = 5470\ J$. The mass of the solution $m_{solution}=40.2\ g+108\ g = 148.2\ g$. Using the formula $q = mc\Delta T$, where $c = 4.18\ J/(g\cdot^{\circ}C)$. Then $\Delta T=\frac{q}{mc}=\frac{5470\ J}{148.2\ g\times4.18\ J/(g\cdot^{\circ}C)}\approx8.8^{\circ}C$. The final temperature $T = 21.5^{\circ}C-8.8^{\circ}C = 12.7^{\circ}C$.
c. Cold - pack, because the reaction is endothermic ($\Delta H^{\circ}=23.7\ kJ/mol>0$) and absorbs heat from the surroundings.