Question

What would be the change in pressure in a sealed 10.0 L vessel due to the formation of N2 gas when the ammonium nitrite in 2.00 L of 1.10 M NH4NO2 decomposes at 25.0°C? -5.3826 atm

Answer 1

First, let's find the number of moles of ammonium nitrite.

1.10 mol/L × 2.00 L = 2.2 mol

The decomposition reaction is

NH₄NO₂ → N₂ + 2 H₂O

Let's calculate the amount of N₂ gas produced.

2.2 mol × 1mol N₂/1mol NH₄NO₂ = 2.2 mol N₂

Then, let's compute P ( pressure) through the ideal gas equation:
P = nRT/ V ( putting value in equation)

n = no. of moles of N₂ = 2.2

R = constant = 0.0821 L-atm/mol-K

T = Temperature = 25°C

= 25+273K = 298 Kelvin

V = volume of vessel = 10 L

P = (2.2)(0.0821L-atm/mol-K)(298 K)/10 L

P = 5.38 atm