R2.3 How far? The extent of chemical change

For ${\text{N}}_2{\text{O}}_4(g)\rightleftharpoons 2{\text{NO}}_2(g)$, the equilibrium expression is: $K_c=\frac{[{\text{NO}}_2{]}^2}{[{\text{N}}_2{\text{O}}_4]}$

$0.060=\frac{x^2}{0.15}$ $x^2=0.060\times 0.15=0.0090$

$x=\sqrt{0.0090}=\mathrm{0.0949...}\approx 0.094{\text{ mol dm}}^{-3}$

The concentration of ${\text{NO}}_2$ at equilibrium is approximately $0.094{\text{ mol dm}}^{-3}$, which matches option B.

We need to find $x=[{\text{NO}}_2]$ using $K_c=\frac{x^2}{0.15}=0.060$, which gives $x^2=0.0090$ and $x\approx 0.094$.

If $x=0.0040$, then $x^2=1.6\times 10^{-5}$, giving $K_c=1.6\times 10^{-5}/0.15\approx 1.1\times 10^{-4}$, which is far too small. Eliminated.

If $x=0.30$, then $x^2=0.090$, giving $K_c=0.090/0.15=0.60$, which is ten times too large. Eliminated.

If $x=0.0090$, then $x^2=8.1\times 10^{-5}$, giving $K_c=8.1\times 10^{-5}/0.15\approx 5.4\times 10^{-4}$, which is still far too small. Eliminated.

Only $x=0.094{\text{ mol dm}}^{-3}$ gives $x^2/0.15=0.0088/0.15\approx 0.059\approx 0.060$. This is the correct answer.

The value of $K_c$ is only altered by a change in temperature. Changes in concentration, pressure, or the addition of a catalyst shift the equilibrium position but leave $K_c$ unchanged.

Since the reaction is exothermic ($\Delta H<0$), heat can be treated as a product. Decreasing the temperature removes heat, shifting the equilibrium to the right to produce more heat (and more ${\text{SO}}_3$).

For an exothermic reaction, decreasing temperature increases the value of $K_c$ because the product-to-reactant ratio at equilibrium increases.

Decreasing the temperature shifts equilibrium right and increases $K_c$. This is the only option that satisfies both conditions in the question.

We need a change that both shifts equilibrium to the right and changes the numerical value of $K_c$. Only temperature can change $K_c$.

Adding more product shifts equilibrium to the left, not the right, and $K_c$ is unchanged. Eliminated.

A catalyst speeds up both forward and reverse reactions equally, reaching equilibrium faster but not shifting the position or changing $K_c$. Eliminated.

Increasing pressure does shift equilibrium to the right (fewer moles of gas on the right: 2 vs 3), but it does not change $K_c$. Eliminated.

Decreasing temperature is the only change that both shifts equilibrium right (for this exothermic reaction) and changes $K_c$ (increases it). This is correct.