Thermodynamics and ThermochemistryhardMCQ SINGLE

Calculate the heat effect produced when a solution of 1 mole of ethanol in 5 moles of water is mixedThermodynamics and Thermochemistry Chemistry Question

Question

Calculate the heat effect produced when a solution of 1 mole of ethanol in 5 moles of water is mixed with a solution of 2 moles of ethanol in 20 moles of water. Given that,<br>$\text{C}_2\text{H}_5\text{OH} + 5\text{H}_2\text{O} \to \text{C}_2\text{H}_5\text{OH (5H}_2\text{O)}; \Delta H = -1120\text{ cal}$<br>$\text{C}_2\text{H}_5\text{OH} + 10\text{H}_2\text{O} \to \text{C}_2\text{H}_5\text{OH (10H}_2\text{O)}; \Delta H = -1760\text{ cal}$<br>$\text{C}_2\text{H}_5\text{OH} + 25/3\text{H}_2\text{O} \to \text{C}_2\text{H}_5\text{OH (25/3H}_2\text{O)}; \Delta H = -1650\text{ cal}$

Answer: B

💡 Solution & Explanation

Initial state: 1 mol in 5 $\text{H}_2\text{O}$ (-1120 cal) + 2 mol in 20 $\text{H}_2\text{O}$ [which is $2 \times (-1760) = -3520\text{ cal}$]. Total initial $H = -4640\text{ cal}$. Final state: 3 mol in 25 $\text{H}_2\text{O}$ [which is $3 \times (-1650) = -4950\text{ cal}$]. Heat effect $\Delta H = H_{\text{final}} - H_{\text{initial}} = -4950 - (-4640) = -310\text{ cal}$. Thus, 310 cal is released.

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