Two solutions, initially at 25°C, were mixed in an insulated bottle. One contained 200 ml of 0.5 M weak monoprotic acid solution. The other contained 80 ml of a solution having 0.5 mole KOH per litre.

Question

Two solutions, initially at 25°C, were mixed in an insulated bottle. One contained 200 ml of 0.5 M weak monoprotic acid solution. The other contained 80 ml of a solution having 0.5 mole KOH per litre. After mixing, the temperature rose to 26°C. Assume that the densities of both the solutions are 1.0 g/ml and that their specific heat capacities are all 1.0 cal/g-K. The amount of heat evolved (in kcal) in the neutralization of 1 mole of the acid is

Answer:

7.00

Solution:

KOH is limiting: 0.080 × 0.5 = 0.04 mol (acid = 0.10 mol, in excess). Heat = 280 × 1.0 × 1 = 280 cal for 0.04 mol neutralized, so per mole = 280/0.04 = 7000 cal = 7 kcal. Answer: 7.

TestHub · Accepted Answer

KOH is limiting: 0.080 × 0.5 = 0.04 mol (acid = 0.10 mol, in excess). Heat = 280 × 1.0 × 1 = 280 cal for 0.04 mol neutralized, so per mole = 280/0.04 = 7000 cal = 7 kcal. Answer: 7.

Chemistry - Thermochemistry Question with Solution | TestHub

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