Question

In: Chemistry

Consider the following data: ΔvapH = 5.56 kJ/mol Specific heat capacity of N2(g) = 29.2 J...

Consider the following data: ΔvapH = 5.56 kJ/mol Specific heat capacity of N2(g) = 29.2 J mol-1 K-1 Specific heat capacity of N2(l) = 56.04 J mol-1 K-1 Boiling point of N2 = 77.355 K What is the energy in the form of heat released when 450 grams of nitrogen gas at 23.0 °C is converted to liquid nitrogen at -200.0 °C? Select one: a. 364 kJ b. 196 kJ c. 392 kJ d. 17.1 kJ

Solutions

Expert Solution

Ti = 23.0 oC

Tf = -200.0 oC

Cg = 29.2 J/mol.oC

boiling point = 77.355 K = 77.355 - 273.15 oC = -195.795 oC

Lets convert mass to mol

Molar mass of N2 = 28.02 g/mol

number of mol

n= mass/molar mass

= 450.0/28.02

= 16.06 mol

Heat released to convert vapour from 23.0 oC to -195.795 oC

Q1 = n*Cg*(Ti-Tf)

= 16.06 mol * 29.2g J/mol.oC *(23--195.795) oC

= 102604.0792 J

Lv = 5.56KJ/mol = 5560J/mol

Lets convert mass to mol

Molar mass of N2 = 28.02 g/mol

number of mol

n= mass/molar mass

= 450.0/28.02

= 16.06 mol

Heat released to convert gas to liquid at -195.795 oC

Q2 = n*Lv

= 16.06 mol *5560 J/mol

= 89293.3619 J

Cl = 56.04 J/mol.oC

Lets convert mass to mol

Molar mass of N2 = 28.02 g/mol

number of mol

n= mass/molar mass

= 450.0/28.02

= 16.06 mol

Heat released to convert liquid from -195.795 oC to -200.0 oC

Q3 = n*Cl*(Ti-Tf)

= 16.06 mol * 56.04l J/mol.oC *(-195.795--200) oC

= 3784.5 J

Total heat released = Q1 + Q2 + Q3

= 102604.0792 J + 89293.3619 J + 3784.5 J

= 195682 J

= 196 KJ

Answer: 196 KJ

queen_honey_blossom answered 2 years ago
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