When 0.49 g of NH4Br was dissolved in 20 g of liquid containing an equimolar amount of KNH2, 0.80 g of ammonia was vaporized. The ΔH for the reaction: NH4+(, l) + NH2-(, l) → 2(l) at 240 K is

Question

When 0.49 g of NH4Br was dissolved in 20 g of liquid $NH_3$ containing an equimolar amount of KNH2, 0.80 g of ammonia was vaporized. The ΔH for the reaction: NH4+($NH_3$, l) + NH2-($NH_3$, l) → 2$NH_3$(l) at 240 K is

Prashant Kotian · Accepted Answer

Correct answer: D. Moles of NH4Br = 0.49 / 98 = 0.005 mol. Heat of neutralization: NH4+ + NH2- → 2. Total ammonia vaporized = 0.80 g. Heat absorbed = 0.80 × 320 = 256 cal. Since heat is absorbed by vaporization, the reaction is exothermic, releasing 256 cal. Enthalpy of reaction per mole = -256 cal / 0.005 mol = -51200 cal/mol = -51.2 kcal/mol.

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