Question

Ammonia under a pressure of 20 atm at $27^o$ C is heated to $327^o$ C in a closed vessel. Under these conditions \[N{H_3}\] is partially decomposed to \[{N_2}\]and \[{H_2}\;\] according to the equation:
After decomposition at constant volume in a vessel, the pressure increases to 45 atm. What is the percentage of ammonia dissociated?

Answer 1

Hint: The gas laws comprise of the five primary laws namely (i) Charles' Law, (ii) Boyle's Law, (iii) Avogadro's Law, (iv) Gay-Lussac Law and (v) Combined Gas Law. These five gas laws invented the relationship between temperature, pressure, volume and the amount of gas. The first gas law i.e. Boyle's Law states that the volume of gas is inversely proportional to the pressure. Charles' Law states that the volume of gas is directly proportional to the temperature. Avogadro's Law states that volume of gas is directly proportional to the amount of gas. When these three gas laws are combined together, it forms the ideal gas law. Gay-Lussac's law states that the pressure of an ideal gas relates directly to its temperature at a constant volume of gas. Combined gas law states that volume of a given amount of gas is proportional to its temperature and pressure.

Complete step by step answer:
Pressure of ammonia (\[N{H_3}\]) at 27°C (i.e. 300 K) = 20 atm (Given)
Let us suppose, Pressure of ammonia (\[N{H_3}\]) at 327°C (i.e. 620 K)= P atm (Assumption)
$PV = nRT$ (Ideal Gas Law)
${P_1}{V_1} = {P_2}{V_2}$ (Boyle’s Law)
$P = kT$ (Gay-Lussac Law)
Applying Gay Lussac Law, find the value of P:
$\dfrac{P}{{620}} = \dfrac{{20}}{{300}}$
$P = 41.3$atm
Let us assume that ‘a’ moles of \[N{H_3}\]are present initially.
It is given that the pressure at equilibrium becomes 45 atm.

(a – 2x) (x) (3x) (At equilibrium)
Total moles at equilibrium= $a - 2x + x + 3x = a + 2x$
We know:
$\dfrac{{{\text{Initial number of moles}}}}{{{\text{Moles at equilibrium}}}} = \dfrac{{{\text{Initial pressure}}}}{{{\text{Equilibrium pressure}}}}$
Putting the values in the above equation, solve for x:
$
  \dfrac{a}{{a + 2x}} = \dfrac{{41.3}}{{45}}
  x = 0.045a
$
Now, we know that amount of \[N{H_3}\]decomposed = 2x, which is equal to:
$2x = 2 \times 0.045a = 0.090a$
Finally, we can calculate the percentage of \[N{H_3}\]decomposed as follows:
$\dfrac{{0.090a}}{a} \times 100 = 9\% $
As a result, the percentage of ammonia dissociated is 9.0%.


Note: Real gases refer to the non-ideal which do not follow the ideal gas law exactly. For real gases, two changes have been incorporated like (i) a constant has been added to the pressure (P) and (ii) a different constant has been subtracted from the volume (V). Thus, the new equation for real gas law is: $(P + a{n^2}) \times (V - nb) = nRT$.