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.

$\text{ Moles of }NH_4NO_2=\text{ Molarity of }NH_4NO_2\times \text{ Volume of }NH_4NO_2$

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

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.

$P=\frac{nRT}{V}=\frac{(2.16mole)\times (0.0821Latm/moleK)\times (298K)}{10L}=5.285atm$

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.