Question

Exercise 14.80 The half-life for the first-order decomposition of nitramide, NH2NO2(aq)→N2O(g)+H2O(l), is 123 min at 15∘C.

Part A If 170 mL of a 0.105 M NH2NO2 solution is allowed to decompose, how long must the reaction proceed to yield 46.0 mL of N2O(g) collected over water at 15∘C and a barometric pressure of 756 mmHg ? (The vapor pressure of water at 15∘C is 12.8 mmHg.) Express your answer using two significant figures. t = min Please show as much relevant work as possible. Thank you

Answer 1

Initial moles of NH2NO2 = molarity x volume in Litres = 0.105 M x 0.17 L = 0.01785 mol

Moles of N2O at 15∘C and a barometric pressure of 756 mmHg:

Given that 46.0 mL of N2O(g) collected over water at 15∘C and a barometric pressure of 756 mmHg

Also given that vapor pressure of water at 15∘C is 12.8 mmHg.

Hence,

Volume of N2O, V = 46 mL = 0.046 L

Pressure of N2O, P = 756 mmHg - vapor pressure of water = 756 mmHg - 12.8 mmHg = 743.2 mmHg

= 743.2/760 atm

= 0.978 atm

temperature T = 15∘C = 15 +273 K= 288 K

Ideal gas equation , PV = n R T , R = 0.0821 L.atm/mol/K

n = PV/ RT

= (0.978 atm) (0.046 L) / (0.0821 L.atm/mol/K x 288 K )

= 0.002 mol

Therefore, moles of N2O = 0.002 mol

Final moles of NH2NO2:

Final moles of NH2NO2 = Initial moles of NH2NO2 - moles of N2O formed

=  0.01785 mol - 0.002 mol

= 0.01585 mol

Calculation of time:

Given that half life = 123 min

[A]o=  0.01785 mol

[A]t = 0.01585 mol

For first order recation,

half life t_1/2 = 0.693 /k where k is rate constant

k = 0.693/ t_1/2 --- Eq (1)

k = 1/t ln { [A]_o/[A]_t} -----Eq (2)

From Eqs (1) and (2),

0.693/ t_1/2 = (1/t) ln {[A]_o/ [A]_t} ------Eq (3)

Substitute all the values in Eq (3),

0.693/ t_1/2 = (1/t) ln {[A]_o/ [A]_t}

0.693/ 123 = (1/t) ln {0.01785/0.01585}

t = 21.1 min