Question

One mole of nitrogen was mixed with 3 moles of hydrogen in a closed 3 litre vessel. \(20 \%\) of nitrogen is converted into \(\mathrm{NH}_{3}\). Then, \(\mathrm{K}_{\mathrm{C}}\) for the \(1 / 2 \mathrm{~N}_{2}+3 / 2 \mathrm{H}_{2}\) \(\rightleftharpoons \mathrm{NH}_{3}\) is(1) \(0.36\) litre mol 1 (2) \(0.46\) litre mol (3) \(0.5 \mathrm{~mol}\) ' litre (4) \(0.2 \mathrm{~mol}^{\text {' }}\) litre

Short answer

The value of \( K_C \) is 0.2 litre mol^{-1}.

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation is as follows: \[ \frac{1}{2} N_2 + \frac{3}{2} H_2 \rightleftharpoons NH_3 \]

Determine initial moles and volume

Initially, we have 1 mole of nitrogen (\(N_2\)) and 3 moles of hydrogen (\(H_2\)) in a 3-litre vessel.

Calculate moles of nitrogen converted

20% of 1 mole of \(N_2\) is converted into \(NH_3\). Therefore, \[ 0.20 \times 1 \text{ mole N}_2 = 0.2 \text{ moles N}_2 \text{ converted} \]

Determine moles of NH_3 formed

Each mole of \(N_2\) forms 2 moles of NH_3. Thus, moles of NH_3 formed are: \[ 0.2 \text{ moles N}_2 \times 2 = 0.4 \text{ moles NH}_3 \]

Calculate moles of nitrogen and hydrogen remaining

After conversion, the moles remaining are: \( \text{Nitrogen: } 1 \text{ mole - } 0.2 \text { mole} = 0.8 \text{ moles} \)\( \text {Hydrogen:} 3 \text{ moles - } 3 \times 0.2 = 2.4 \text{ moles} \)

Calculate equilibrium concentrations

Divide the moles by the volume to get the concentrations: \[ [N_2] = \frac{0.8}{3} \text{ mol/L} = 0.267 \text{ mol/L} \] \[ [H_2] = \frac{2.4}{3} \text{ mol/L} = 0.8 \text{ mol/L} \] \[ [NH_3] = \frac{0.4}{3} \text{ mol/L} = 0.133 \text{ mol/L} \]

Write the expression for the equilibrium constant K_C

The expression for the equilibrium constant is: \[ K_C = \frac{[NH_3]}{[N_2]^{1/2} [H_2]^{3/2}} \]

Substitute values and calculate K_C

Substitute the concentrations into the expression to calculate \( K_C \): \[ K_C = \frac{0.133}{(0.267)^{1/2} (0.8)^{3/2}} \]Simplifying this gives: \[ K_C = \frac{0.133}{\sqrt{0.267} \times 0.8^{1.5}} \] \[ K_C = 0.2 \text{ litre mol}^{-1} \]

Key concepts

chemical equilibrium

In a chemical reaction, chemical equilibrium is the point at which both reactants and products are present in concentrations which have no further tendency to change with time. This state occurs when the forward reaction proceeds at the same rate as the reverse reaction. At equilibrium, the amounts of reactants and products remain constant.

• In our example, the reaction is: \(\frac{1}{2} N_2 + \frac{3}{2} H_2 \rightleftharpoons NH_3\)
  
• When we establish equilibrium, it means the rate at which \(N_2 \) and \(H_2\) are converting to \(NH_3\) is equal to the rate \(NH_3\) is decomposing back to \(N_2 \) and \(H_2\).
  

reaction quotient

The reaction quotient, \(Q_c\), is a ratio that relates the concentrations of products and reactants at any point during a reaction. It is expressed using the same formula as the equilibrium constant, \(K_c\), but measures the system at any point before reaching equilibrium.

• For our example, \(Q_c\) can be calculated using: \[ Q_c = \frac{[NH_3]}{[N_2]^{1/2} [H_2]^{3/2}} \]
  
• If \(Q_c < K_c\), the reaction will move forward to form more products until equilibrium is achieved.
  
• If \(Q_c > K_c\), the reaction will move in the reverse direction until equilibrium is reached.

Calculating \(Q_c\) for initial setups can help us understand in which direction the reaction will proceed to reach equilibrium.

mole conversion

Mole conversion is a method of converting the quantity of a substance in moles to its equivalent in terms of molecules, other reactants, or products using the stoichiometric coefficients from the balanced chemical equation.

• In the given exercise, we see 20% of \(N_2\) gets converted.
  
• This means \(0.2\) moles of \(N_2\) gets converted, producing \(0.4\) moles of \(NH_3\) due to the mole ratio in the balanced equation.
  
• \