Answer to Question #179409 in Chemistry for dtr

Question

A 22.11 g sample of liquid nitrogen (N₂) starts at a temperature of -288 ^oC. How much energy is required to heat it to its boiling point of -196 ^oC and boil it completely?

s = 1.04 J/g^oC delta H_v = 2.79 kJ/mol

Answer 1

Solution:

The molar mass of N₂ is 28.013 g/mol.

Hence,

Moles of N₂ = (22.11 g N₂) × (1 mol N₂ / 28.013 g N₂) = 0.789 mol N₂

n(N₂) = 0.789 mol

This problem can be summarized thusly:

1) Sample of liquid nitrogen (N₂) is heated from -288°C to -196°C.

The heat is calculated by using the specific heat of liquid nitrogen (s = 1.04 J/g°C) and the equation:

q = m × S × ΔT

q₁ = 22.11 g × 1.04 J/g°C × (-196 - (-288))°C = 2115.5 J = 2.116 kJ

q₁ = 2.116 kJ

2) Liquid nitrogen is vaporized at -196°C.

The heat is calculated by multiplying the moles of liquid nitrogen by the heat of vaporization (ΔH_v).

q₂ = n(N₂) × ΔH_v = 0.789 mol × 2.79 kJ/mol = 2.201 kJ

q₂ = 2.201 kJ

q = q₁ + q₂ = 2.116 kJ + 2.201 kJ = 4.317 kJ = 4.32 kJ

q = 4.32 kJ

Answer: 4.32 kJ of energy is required.