Question

Nitric oxide (NO) reacts with molecular oxygen as follows: 2NO(g) + O2(g) → 2NO2(g) Initially NO and O2 are separated in two different chambers connected by a valve. When the valve is opened, the reaction quickly goes to completion. Determine what gases remain at the end and calculate their partial pressures. Assume that the temperature remains constant at 25°C. Initial conditions are as follows: NO: 3.99 L, 0.500 atm O2: 1.97 L, 1.00 atm

PNO = ???atm

PO2 = ???atm

PNO2 = ???atm

Answer 1

Answer -

Given things are,

T = 25^oC = 298.15^o K

V(NO) = 3.99L

V(O₂) = 1.97L

P(NO) = 0.500atm

P(O₂) = 1.00atm

First we use ideal gas law to find out the mles of each gas PV=nRT

Here R = 0.0821Latm/molK

T = 298.15^oK

For NO,

n(NO) = PV/RT

n(NO) = 0.500 * 3.99 / (0.0821*298.15)

n(NO) = 0.0815 mole NO

For O₂ ,

n(O₂) = 1.00* 1.97 /(0.0821*298.15)

n(O₂) = 0.0804 mole O₂

According to the reaction mole ratio is 2:1

Nitrogen oxide disappears leaving remaining oxygen besides nitrogen dioxide formed.

The mole gas number will be

mole of NO = 0.0815 - 0.0815(2/2) = 0 mole

mole of O₂ = 0.0815 - 0.0804 (1/2) = 0.0413 mole O₂

mole of NO₂ = 0.0815*(2/2) = 0.0815 mole NO2

Again we use ideal gas law to find out the answer,

Here we use total volume which is 3.99L + 1.97L

P(NO) = 0 atm

P(O2) = 0.0413 *0.0821 * 298.15 /(3.99+1.97)

= 0.169 atm

P(NO2) = 0.0815 * 0.0821*0298.15 / (3.99+1.99)

= 0.335atm