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 2.40 l of 0.900 m nh4no2 decomposes at 25.0°c?

Answer 1

Answer : The pressure of the gas will be, 5.285 atm

Solution : Given,

Volume of
`N_2` gas = 10 L

Temperature of gas =
`25^oC=273+25=298K`
`(0^oC=273K)`

Volume of
`NH_4NO_2` = 2.40 L

Molarity of the ammonium nitrate solution = 0.9 M

First we have to calculate the moles of ammonium nitrate.

`\text{ Moles of }NH_4NO_2=\text{ Molarity of }NH_4NO_2* \text{ Volume of }NH_4NO_2`

`\text{ Moles of }NH_4NO_2=(0.9mole/L)* (2.40L)=2.16moles`

The balanced chemical reaction is,

`NH_4NO_2\rightarrow N_2+2H_2O`

From the balanced reaction, we conclude that

1 mole of
`NH_4NO_2` decomposes to give 1 mole of
`N_2` gas

2.16 moles of
`NH_4NO_2` decomposes to give 2.16 moles of
`N_2` gas

Now we have to calculate the pressure of the gas.

using ideal gas equation,

`PV=nRT`

where,

P = pressure of the gas

V = volume of the gas

T= temperature of the gas

n = number of moles of gas

R = gas constant = 0.0821 Latm/moleK

Now put all the given values in ideal gas equation, we get pressure of the gas.

`P=(nRT)/(V)=((2.16mole)* (0.0821Latm/moleK)* (298K))/(10L)=5.285atm`

Therefore, the pressure of the gas will be, 5.285 atm

Answer 2

The change in pressure in a sealed 10.0L vessel is 5.28 atm

calculation

The pressure is calculated using the ideal gas equation

That is P=n RT

where;

P (pressure)= ?

v( volume) = 10.0 L

n( number of moles) which is calculated as below

write the equation for decomposition of NH₄NO₂

NH₄NO₂ → N₂ +2H₂O

Find the moles of NH₄NO₂

moles = molarity x volume in liters

= 2.40 l x 0.900 M =2.16 moles

Use the mole ratio to determine the moles of N₂

that is from equation above NH₄NO₂:N₂ is 1:1 therefore the moles of N₂ is also =2.16 moles

R(gas constant) =0.0821 l.atm/mol.K

T(temperature) = 25° c into kelvin = 25 +273 =298 K

make p the subject of the formula by diving both side by V

P = nRT/V

p ={ (2.16 moles x 0.0821 L.atm/mol.K x 298 K) /10.0 L} = 5.28 atm.