Question

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).

The molar heat of solution of NH4Br in liquid NH3 at this concentration is

• A. +80.0 cal
• B.
  • 80.0 cal
• C.
  • 8.0 kcal
• D. +8.0 kcal

Answer 1

Answer: (C)

• 8.0 kcal

Answer 2

Here's the step-by-step solution:

1. Calculate the moles of NH4Br dissolved: The molar mass of NH4Br (N=14, H=1, Br=80) is $14 + 4(1) + 80 = 98 \text{ g/mol}$. Moles of NH4Br = $\frac{\text{Mass of NH4Br}}{\text{Molar mass of NH4Br}} = \frac{0.98 \text{ g}}{98 \text{ g/mol}} = 0.01 \text{ mol}$.

2. Calculate the total heat released by the dissolution process: The heat released by the dissolution of NH4Br causes the vaporization of liquid NH3. Heat released = Mass of NH3 vaporized $\times$ Heat of vaporization of NH3 Heat released = $0.25 \text{ g} \times 320.0 \text{ cal/g} = 80.0 \text{ cal}$. Since heat is released, the enthalpy change ($\Delta H$) for the dissolution of 0.01 mol of NH4Br is $-80.0 \text{ cal}$.

3. Calculate the molar heat of solution: Molar heat of solution = $\frac{\text{Heat released}}{\text{Moles of NH4Br}} = \frac{-80.0 \text{ cal}}{0.01 \text{ mol}} = -8000 \text{ cal/mol}$. Converting to kilocalories: $-8000 \text{ cal/mol} = -8.0 \text{ kcal/mol}$.