Let us solve the problem step by step.

(i) $K_c$ expression

The chemical equation is:

$2CO(g) + 2NO(g) \rightleftharpoons 2CO_2(g) + N_2(g)$

The equilibrium constant expression for this reaction is:

[ K_c = \frac{[CO_2]^2 [N_2]}{[CO]^2 [NO]^2} ]

(ii) Units of $K_c$

For a reaction where the stoichiometric coefficients in the balanced equation sum to different numbers for reactants and products:

• In this case: \text{Number of moles of products = 3 (2 from CO_2$+ 1 from$N_2)}. \text{Number of moles of reactants = 4 (2 from CO$+ 2 from$NO)}.

  The difference in moles: $\Delta n = \text{moles of products - moles of reactants} = 3 - 4 = -1$

Thus, the units of $K_c$ are:

$\text{Units of } K_c = \text{(mol/dm}^3\text{)}^{\Delta n}$

Since $\Delta n = -1$:

$\text{Units of } K_c = \frac{1}{\text{mol/dm}^3} = \text{dm}^3 \text{mol}^{-1}$

(iii) Determining the value of $K_c$

Given:

• Initial moles of $CO$ = $0.45 \, \text{mol}$
• Initial moles of $NO$ = $0.45 \, \text{mol}$
• At equilibrium, moles of $NO$ = $0.25 \, \text{mol}$
• Total volume $V = 1.0 \, \text{dm}^3$.

Step 1: Change in moles of $NO$

The change in moles of $NO$ is:

$\Delta [NO] = 0.45 - 0.25 = 0.20 \, \text{mol}$

From the stoichiometry of the reaction:

$\text{For every 2 moles of } NO \text{ reacted, 2 moles of } CO \text{ react, and } 2 \text{ moles of } CO_2 \text{ are formed.}$

Thus: [ \Delta [CO] = \Delta [NO] = 0.20 , \text{mol} ] [ \Delta [CO_2] = \frac{2}{2} \cdot \Delta [NO] = 0.20 , \text{mol} ] [ \Delta [N_2] = \frac{1}{2} \cdot \Delta [NO] = 0.10 , \text{mol} ]

Step 2: Equilibrium concentrations

• $[CO]_\text{eq} = 0.45 - 0.20 = 0.25 \, \text{mol} \, \text{dm}^{-3}$
• $[NO]_\text{eq} = 0.25 \, \text{mol} \, \text{dm}^{-3}$
• $[CO_2]_\text{eq} = 0 + 0.20 = 0.20 \, \text{mol} \, \text{dm}^{-3}$
• $[N_2]_\text{eq} = 0 + 0.10 = 0.10 \, \text{mol} \, \text{dm}^{-3}$

Step 3: Substituting into $K_c$

[ K_c = \frac{[CO_2]^2 [N_2]}{[CO]^2 [NO]^2} ]

Substitute the values: $K_c = \frac{(0.20)^2 (0.10)}{(0.25)^2 (0.25)^2}$

Calculate step by step:

• Numerator: $(0.20)^2 (0.10) = 0.04 \cdot 0.10 = 0.004$
• Denominator: $(0.25)^2 (0.25)^2 = (0.0625) (0.0625) = 0.00390625$

Thus: $K_c = \frac{0.004}{0.00390625} \approx 1.02$

Final Answers:

1. $K_c = \frac{[CO_2]^2 [N_2]}{[CO]^2 [NO]^2}$
2. Units of $K_c$: $\text{dm}^3 \text{mol}^{-1}$
3. Value of $K_c$: $\mathbf{1.02}$

Would you like a further breakdown or clarification?

Follow-Up Questions:

1. How would $K_c$ change if the temperature were increased for this exothermic reaction?
2. What is the significance of the units of $K_c$ in determining the reaction's equilibrium?
3. How can $K_c$ be used to determine the direction of a reaction?
4. What assumptions are made when calculating $K_c$?
5. How would the equilibrium concentrations change if the volume of the system doubled?

Tip: Always verify your equilibrium concentrations carefully to ensure accurate calculations of $K_c$.