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 1.40 L of 1.40 M NH4NO2 decomposes at 25.0°C?

Answer 1

Answer:

3.80 atm

Explanation:

First of all, ammonium nitrite decomposes to produce nitrogen gas and liquid water at the given room temperature:

$NH_4NO_2 (s)\rightarrow N_2 (g) + 2 H_2O (l)$

Notice that the solid decomposes fully and produces the same number of moles of nitrogen:

$n_{NH_4NO_2} = n_{N_2}$

Find moles of ammonium nitrite multiplying its molarity by its volume:

$n_{NH_4NO_2} = c_{NH_4NO_2} V_{NH_4NO_2}$

Express moles of nitrogen in terms of the ideal gas law:

$pV_{N_2}=n_{N_2}RT \therefore n_{N_2} = \frac{pV_{N_2}}{RT}$

Substitute the two expressions above into the molar ratio equation:

$c_{NH_4NO_2} V_{NH_4NO_2}=\frac{pV_{N_2}}{RT}$

Solve for the pressure:

$p = \frac{c_{NH_4NO_2} V_{NH_4NO_2}RT}{V_{N_2}}=\frac{1.40 M\cdot 1.40 L\cdot 0.08206 \frac{L atm}{mol K}\cdot 298.15 K}{10.0 L}=4.80 atm$

Notice that here R is the ideal gas law constant and we've converted the temperature into the absolute temperature: $T = 25.0^oC + 273.15 K$.

Therefore, the overall final pressure would be 4.80 atm. Since initially we had air in the vessel at standard conditions (1.00 atm), the change in pressure would be 4.80 atm - 1.00 atm = 3.80 atm.