A solution of 500 ml of 2 M- is added to 500 ml of 2 M- and the mixture is well shaken. The rise in — Thermodynamics and Thermochemistry Chemistry Question
Question
A solution of 500 ml of 2 M-$KOH$ is added to 500 ml of 2 M-$HCl$ and the mixture is well shaken. The rise in temperature T_1 is noted. The experiment is again performed using 250 ml of $KOH$ solution and 500 ml of $HCl$ solution and rise in temperature T_2 is again noted. Which of the following is correct?
💡 Solution & Explanation
In Exp 1: 500 mL of 2 M $KOH$ (1.0 mol OH-) + 500 mL of 2 M $HCl$ (1.0 mol H+). Heat released q1 = 1.0 * 57.1 kJ. Total volume V1 = 1000 mL. Rise in temperature T1 = q1 / (V1 * s) = 57.1 / (1000 * s). In Exp 2: 250 mL of 2 M $KOH$ (0.5 mol OH-) + 500 mL of 2 M $HCl$ (1.0 mol H+). Since $KOH$ is limiting, 0.5 mol of $H_2O$ is formed. Heat released q2 = 0.5 * 57.1 kJ. Total volume V2 = 750 mL. Rise in temperature T2 = q2 / (V2 * s) = 28.55 / (750 * s). Ratio T1/T2 = (57.1 / 1000) / (28.55 / 750) = 1.5. Thus, T1 is 1.5 times as large as T2.