Chemistry - Thermochemistry Question with Solution | TestHub
The thermal effects of reactions in liquid NH3 at -33°C were measured by observing the quantity of liquid NH3 vaporized by the process of interest. The heat of vaporization of NH3 at -33°C is 320.0 cal/g. When 0.98 g of NH4Br was dissolved in 20 g of liquid NH3, 0.25 g of NH3 was vaporized. (Br = 80)
When 0.49 g NH4Br was dissolved in 20 g of liquid NH3 containing an equimolar amount of KNH2, 0.80 g of ammonia was vaporized. The ΔH for the reaction:
NH4+(NH3, l) + NH2-(NH3, l) → 2NH3(l)
at 240 K is
Options:
Answer:
Solution:
First, calculate the moles of NH₄Br:
0.49 g NH4Br = 0.49/98 = 0.00500 mol
This means that both NH₄⁺ and NH₂⁻ are present in 0.00500 mol amounts.
Next, calculate the heat absorbed by vaporizing 0.80 g of NH₃:
Heat absorbed = 0.80 g × 320 cal/g = 256 cal
This heat is released by the reaction. To find the heat released per mole of reaction, divide the total heat released by the moles of reactant:
Heat per mole = (-256 cal)/0.00500 mol = -51,200 cal/mol = -51.2 kcal/mol
The answer is (d).
