Question

Two bulbs are connected by a stopcock. The large bulb, with a volume of 6.00 L, contains nitric oxide at a pressure of 0.850 atm, and the small bulb, with a volume of 1.50 L, contains oxygen at a pressure of 2.50 atm. The temperature at the beginning and the end of the experiment is 22∘C .
After the stopcock is opened, the gases mix and react.

2NO(g)+O2(g)→2NO2(g)

1. Which gases are present at the end of the experiment?
2. What are the partial pressures of the gases? If the gas was consumed completely, put 0 for the answer.

Answer 1

Answer:

(I). The gases are present at the end of the experiment  are O₂ and NO₂

(II).  The pressure of O₂ is 0.158 atm.

The pressure of NO₂ is 0.681 atm

The Pressure of NO is zero.

Explanation:

Given that,

Volume of large bulb = 6.00 L

Pressure = 0.850 atm

Volume of small bulb = 1.50 L

Pressure = 2.50 atm

Temperature = 22°C = 295 K

We need to calculate the moles in NO

Using formula of moles

$n=\dfrac{PV}{RT}$

Put the value into the formula

$n=\dfrac{0.850\times6.00}{0.0821\times295}$

$n=0.211\ moles$

We need to calculate the moles in O

Using formula of moles

$n=\dfrac{PV}{RT}$

Put the value into the formula

$n=\dfrac{2.50\times1.50}{0.0821\times295}$

$n=0.155\ moles$

The balance equation for the reaction is

$2NO+O_{2}\Rightarrow 2NO_{2}$

So, The gases are present at the end of the experiment  are O₂ and NO₂

We need to calculate the remaining moles of O₂

Using formula for remaining moles

Moles of  O₂ remaining = $0.155-\dfrac{0.211}{2}$

$Moles\ of \ O_{2}\ remaining =0.049\ moles$

Moles of NO₂ = 0.211 moles

Total volume $V=V_{l}+V_{s}$

Put the value into the formula

$V=6.00+1.50$

$V=7.5\ V$

(2). If the gas was consumed completely

We need to calculate the pressure of O₂

Using formula of pressure

$P=\dfrac{moles\times R\times T}{V}$

Put the value into the formula

$P=\dfrac{0.049\times0.0821\times295}{7.5}$

$P=0.158\ atm$

We need to calculate the pressure of NO₂

Using formula of pressure

$P=\dfrac{moles\times R\times T}{V}$

Put the value into the formula

$P=\dfrac{0.211\times0.0821\times295}{7.5}$

$P=0.681\ atm$

If the gas was consumed completely

Then, Pressure of NO is zero.

Hence, (I). The gases are present at the end of the experiment  are O₂ and NO₂

(II).  The pressure of O₂ is 0.158 atm.

The pressure of NO₂ is 0.681 atm

The Pressure of NO is zero.