Question

The heat of vaporization AHv, of ammonia (NH3) is 23.4 kJ/mol. Calculate the change in entropy AS when 23. g of ammonia boils at -33.5 °C.

Be sure your answer contains a unit symbol and the correct number of significant digits.

Answer 1

The change in entropy when 23 g of ammonia boils at -33.5 °C is 0.178 kJ/K. To find this, the mass of ammonia was converted to moles, the given temperature was converted to Kelvin, and then the heat of vaporization was divided by the temperature.

Step-by-step explanation:

To calculate the change in entropy (ΔS) when 23 g of ammonia boils at -33.5 °C, we'll use the heat of vaporization (ΔHv) of ammonia, which is 23.4 kJ/mol. First, we convert the given mass of ammonia to moles by using its molar mass (17.03 g/mol for NH3). This gives us:

23 g / 17.03 g/mol = 1.351 moles of ammonia.

Next, we use the formula ΔS = ΔHv / T, where T is the temperature in kelvin.

Since the boiling occurs at -33.5 °C, we first need to convert this temperature to Kelvin:

T = -33.5 °C + 273.15 = 239.65 K.

The total change in entropy for the vaporization of ammonia is then:

ΔS = (23.4 kJ/mol × 1.351 mol) / 239.65 K,

ΔS = 31.61 kJ / 239.65 K = 0.1318 kJ/K × 1.351 = 0.178 kJ/K (or 178 J/K).

This is the change in entropy for the given amount of ammonia with the correct number of significant digits and units.

Answer 2

ΔS(23g NH₃) = 132.2 j/K ≅ 130 j/K (~2 sig. figs. based on 2 sig. figs. in 23 grams NH₃).

Step-by-step explanation:

Given ΔHv(NH₃) = 23.4 Kj/mol = 23,400 joules/mole

T = -33.5°C = (-33.5 + 273)K = 239.5K

molar ΔS (at boiling) = ΔHv/T = 23,400 j/mol / 239.5K = 97.70 j/mol·K

ΔS(23g NH₃) = molar ΔS (at boiling) x (moles NH₃) = 97.70j/mol·K x (23g/17g·mol⁻¹) = 132.2 j/K ≅ 130 j/K (~2 sig. figs. based on 2 sig. figs. in 23 grams NH₃).