# IB DP Chemistry Reactivity 2.3 How far? The extent of chemical change IB Style Question Bank HL Paper 1

### Question

This reaction has an equilibrium constant $$K_{\mathrm{c}}=650$$ at a certain temperature.

$\mathrm{NO}_2(\mathrm{~g})+\mathrm{SO}_2(\mathrm{~g}) \rightleftharpoons \mathrm{NO}+\mathrm{SO}_3(\mathrm{~g})$

What is the equilibrium constant for the following reaction at the same temperature?

$\frac{1}{2} \mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{SO}_3(\mathrm{~g}) \rightleftharpoons \frac{1}{2} \mathrm{NO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{SO}_2(\mathrm{~g})$

A. $$\sqrt{650}$$

B. $$\frac{1}{650}$$

C. $$\frac{1}{\sqrt{650}}$$

D. $$\frac{1}{2} \times 650$$

Markscheme: C

The equilibrium constant ($$K_c$$) is related to the balanced chemical equation and the concentrations of the reactants and products at equilibrium. For the given reaction:

$\mathrm{NO}_2(\mathrm{g}) + \mathrm{SO}_2(\mathrm{g}) \rightleftharpoons \mathrm{NO} + \mathrm{SO}_3(\mathrm{g})$

The equilibrium constant is given as $$K_c = 650$$.

Now, consider the second reaction:

$\frac{1}{2} \mathrm{NO}(\mathrm{g}) + \frac{1}{2} \mathrm{SO}_3(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{NO}_2(\mathrm{g}) + \frac{1}{2} \mathrm{SO}_2(\mathrm{g})$

The equilibrium constant ($$K_{c2}$$) for the reversed reaction is related to the original equilibrium constant ($$K_c$$) and multiplied by $\frac{1}{2}$ by the following expression:

$K_{c2} = \frac{1}{\sqrt{K_c}}$

C. $$\frac{1}{\sqrt{650}}$$

### Question

Which equilibrium constant corresponds to the spontaneous reaction with the most negative value of $$ΔG^{\theta }$$?
A. $$4.9 × 10^{-3}$$
B. $$8.2 × 10^{-3}$$
C. $$4.9 × 10^2$$
D. $$8.2 × 10^2$$

Markscheme: D

The relationship between the Gibbs free energy change ($$ΔG^{\theta}$$) and the equilibrium constant ($$K$$) at a given temperature is given by the equation:

$ΔG^{\theta} = -RT \ln K$

where:
$$R$$ is the gas constant (8.314 J/(mol·K)),
$$T$$ is the absolute temperature in Kelvin,
$$K$$ is the equilibrium constant.

Since $$\Delta G^{\theta}$$ is related to the natural logarithm of $$K$$, a more negative value of $$ΔG^{\theta}$$ corresponds to a larger value of $$K$$.

Therefore, the equilibrium constant corresponding to the most negative value of $$ΔG^{\theta}$$ is the largest value among the given options. In this case, that would be:

D. $$8.2 × 10^2$$

### Question

Which is correct when ∆H – T∆S = 0?

A. Forward reaction is favoured.
B. Reverse reaction is favoured.
C. Reaction is in a state of equilibrium.
D. No chemical changes can occur.

Markscheme: C

The expression $$\Delta H – T \Delta S = 0$$ is related to the Gibbs free energy change ($$\Delta G$$) for a reaction and is represented by the equation:

$\Delta G = \Delta H – T \Delta S$

When $$\Delta G = 0$$, it means that the reaction is in a state of equilibrium. At equilibrium, the forward and reverse reactions occur at the same rate, and there is no net change in the concentrations of reactants and products.

C. Reaction is in a state of equilibrium.

### Question

Which condition will cause the given equilibrium to shift to the right?

$$Ag^+(aq) + Cl^-(aq) \rightleftharpoons AgCl(s)$$

A. One half of solid AgCl is removed.
D. The system is subjected to increased pressure.

Markscheme: C

To predict the effect on the equilibrium position, we can use Le Chatelier’s principle, which states that if a system at equilibrium is subjected to a change, the system will adjust itself to counteract that change.

In the given equilibrium:

$\text{Ag}^+(aq) + \text{Cl}^-(aq) \rightleftharpoons \text{AgCl}(s)$

Let’s consider each option:

A. One half of solid AgCl is removed: This is equivalent to removing a product. According to Le Chatelier’s principle, the system will shift to the right to counteract the loss of product. Therefore, the equilibrium will shift to the right.

B. Water is added: Adding water doesn’t affect the concentrations of the ions or the solid. It’s essentially a change in volume, but it won’t shift the equilibrium to the right.

C. Solid NaCl is added: Since NaCl dissociates into Na⁺ and Cl⁻ ions in the solution, adding solid NaCl introduces more Cl⁻ ions. According to Le Chatelier’s principle, the system will shift to the right to counteract the increase in Cl⁻ ions. Therefore, the equilibrium will shift to the right.

D. The system is subjected to increased pressure: This system involves only aqueous ions and a solid, so pressure changes won’t have a significant effect on the equilibrium position. It’s not a factor that would cause a shift to the right.

### Question

Which values of equilibrium constant, K, and Gibbs free energy, ∆G, favour the reverse reaction of an equilibrium?

Markscheme: C

The relationship between the equilibrium constant ($$K$$) and the Gibbs free energy change ($$\Delta G$$) is given by the equation:

$\Delta G = -RT \ln K$

From the table, we can make the following observations:

1. $$K > 1$$ and $$\Delta G > 0$$: This corresponds to a non-spontaneous reaction favoring the reverse direction.

2. $$K > 1$$ and $$\Delta G < 0$$: This corresponds to a spontaneous reaction favoring the forward direction.

3. $$K < 1$$ and $$\Delta G > 0$$: This corresponds to a spontaneous reaction favoring the reverse direction.

4. $$K < 1$$ and $$\Delta G < 0$$: This corresponds to a non-spontaneous reaction favoring the forward direction.

Now, let’s compare these observations with the options provided:

Ans is C.

Scroll to Top