Question

In some aquatic ecosystems, nitrate (NO3-) is converted to nitrite (NO2), which then decomposes to nitrogen and water. As an example of this second reaction, consider the decomposition of ammonium nitrite: NH4NO2 (aq) →→N (8) +2H2O(1) 2nd attempt Feedback W See Periodic Table See Hint 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 1.00 L of 1.10 M NH4NO2 decomposes at 25.0°C? ® 6.73 atm 1st attempt Feedback M See Periodic Table See Hint 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 1.00 L of 1.10 M NH4NO2 decomposes at 25.0°C? * 2.69 atm

Answer 1

Reaction:

NH4NO2(aq) $\rightarrow$ N2(g) + 2 H2O (l)

from the reaction 1 mole of NH4NO2 produces 1 mole of N2 and 2 moles of H2O.

OR, nNH4NO2 = nN2 =2* nH2O

Given:

Molarity of NH4NO2, Ms = 1.10 M

Volume of solution, Vs = 1 L

Calculate moles of NH4NO2,

number of moles of NH4NO2, ns = molarity * volume

ns = 1.10*1 = 1.1 moles

$\therefore$ 1.1 moles of NH4NO2 will produces 1.1 mole of N2 and 2.2 moles of H2O.

nN2 = 1.1 moles

Volume of vessel = 10 L

volume of solution = 1 L

$\therefore$ volume occupied by N2 gas = 10-1.0 L = 9.0 L

using ideal gas law to calculate pressure of N2 formed:

PV = nRT

where P = pressure of N2, V = volume occupied by N2 = 9 L, n = moles of N2 formed = 1.1 moles, R = gas constant = 0.0821 L.atm/K/mol, T = temperature = 25oC = (273+25) K =298 K

substituting the values in above formula we get:

P* 9 = 1.1 * 0.0821*298

or P = 1.1 * 0.0821*298/8.9604

P = 2.99

P $\approx$ 3.00 atm

Therefore change in pressure due to N2 formation is 3 atm.

*Note: amount of water produced in the reaction is (2.2*18 =)39.6 g or 39.6 mL (assuming density = 1g/mL) is small compared to solution volume of 1 L and thus will not effect much to the calculations. The result will still be the same 3 atm.