IB DP Chemistry 17.1 The equilibrium law HL Paper 1

Question

The indicator, HIn is used in a titration between an acid and base. Which statement about the dissociation of the indicator, HIn is correct?

\[{\text{HIn(aq)}} \rightleftharpoons {{\text{H}}^ + }{\text{(aq)}} + {\text{I}}{{\text{n}}^ – }{\text{(aq)}}\]

colour A                             colour B

A.     In a strongly alkaline solution, colour B would be observed.

B.     In a strongly acidic solution, colour B would be observed.

C.     \({\text{[I}}{{\text{n}}^ – }{\text{]}}\) is greater than [HIn] at the equivalence point.

D.     In a weakly acidic solution colour B would be observed.

▶️Answer/Explanation

A

Option A: In a strongly alkaline solution hydroxide ions will absorb H+ ions, hence equilibrium will shift towards forward direction and colour B of Inwill be observed here. 

Question

When gaseous nitrosyl chloride, NOCl (g), decomposes, the following equilibrium is established:

\[{\text{2NOCl(g)}} \rightleftharpoons {\text{2NO(g)}} + {\text{C}}{{\text{l}}_2}{\text{(g)}}\]

2.0 mol of NOCl(g) were placed in a \({\text{1.0 d}}{{\text{m}}^{\text{3}}}\) container and allowed to reach equilibrium. At equilibrium 1.0 mol of NOCl(g) was present. What is the value of \({K_{\text{c}}}\)?

A.     0.50

B.     1.0

C.     1.5

D.     2.0

▶️Answer/Explanation

A

\[{\text{2NOCl(g)}} \rightleftharpoons {\text{2NO(g)}} + {\text{C}}{{\text{l}}_2}{\text{(g)}}\]

Initially 2 moles of NOCl were there. Since at equilibrium 1 moles of NOCl are remaining it means 1 mol of NOCl has reacted, which would produce 1 mol NO and 0.5 moles Cl2 at equilibrium. 

Kc = [NO]2[Cl2]/[NOCl]2 

Volume is 1 dm3.

Kc = [1]2[0.5]/[1]2 = 0.5

Question

What is the relationship between \({\text{p}}{K_{\text{a}}}\), \({\text{p}}{K_{\text{b}}}\) and \({\text{p}}{K_{\text{w}}}\) for a conjugate acid–base pair?

A.     \({\text{p}}{K_{\text{a}}} = {\text{p}}{K_{\text{w}}} + {\text{p}}{K_{\text{b}}}\)

B.     \({\text{p}}{K_{\text{a}}} = {\text{p}}{K_{\text{w}}} – {\text{p}}{K_{\text{b}}}\)

C.     \({\text{p}}{K_{\text{a}}} \times {\text{p}}{K_{\text{b}}} = {\text{p}}{K_{\text{w}}}\)

D.     \(\frac{{{\text{p}}{K_{\text{a}}}}}{{{\text{p}}{K_{\text{b}}}}} = {\text{p}}{K_{\text{w}}}\)

▶️Answer/Explanation

B

\({\text{p}}{K_{\text{a}}} +{\text{p}}{K_{\text{b}}} = {\text{p}}{K_{\text{w}}}\)

So, \({\text{p}}{K_{\text{a}}} = {\text{p}}{K_{\text{w}}} – {\text{p}}{K_{\text{b}}}\)

Question

A mixture of 2.0 mol of \({{\text{H}}_{\text{2}}}\) and 2.0 mol of \({{\text{I}}_{\text{2}}}\) is allowed to reach equilibrium in the gaseous state at a certain temperature in a \({\text{1.0 d}}{{\text{m}}^{\text{3}}}\) flask. At equilibrium, 3.0 mol of HI are present. What is the value of \({K_{\text{c}}}\) for this reaction?

\[{{\text{H}}_{\text{2}}}{\text{(g)}} + {{\text{I}}_{\text{2}}}{\text{(g)}} \rightleftharpoons {\text{2HI(g)}}\]

A.     \({K_{\text{c}}} = \frac{{{{(3.0)}^2}}}{{{{(0.5)}^2}}}\)

B.     \({K_{\text{c}}} = \frac{{3.0}}{{{{(0.5)}^2}}}\)

C.     \({K_{\text{c}}} = \frac{{{{(3.0)}^2}}}{{{{(2.0)}^2}}}\)

D.     \({K_{\text{c}}} = \frac{{{{(0.5)}^2}}}{{{{(3.0)}^2}}}\)

▶️Answer/Explanation

A

At equilibrium 3 moles of HI are formed which means 1.5 moles of each H2 and I2 are consumed. 

Initially, 2 moles each of H2 and I2 were there. So, at equilibrium, o.5 moles each of H2 and I2 are remaining. Volume = 1 dm3

Kc = [HI]2/[H2][I2]

\({K_{\text{c}}} = \frac{{{{(3.0)}^2}}}{{{{(0.5)}^2}}}\)

Question

The equation for the reaction between two gases, A and B, is:

\[{\text{2A(g)}} + {\text{3B(g)}} \rightleftharpoons {\text{C(g)}} + {\text{3D(g)}}\]

When the reaction is at equilibrium at 600 K the concentrations of A, B, C and D are 2, 1, 3 and 2 \({\text{mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) respectively. What is the value of the equilibrium constant at 600 K?

A.     \(\frac{1}{6}\)

B.     \(\frac{9}{7}\)

C.     3

D.     6

▶️Answer/Explanation

D

Kc = [C][D]3/[A]2[B]3

Kc = [3][2]3/[2]2[1]3

Kc = 6

Scroll to Top