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A-Level CAIE Chemistry A223.3 Entropy change, ΔSQuestion Bank

(a)

Under conditions of high pressure and a catalyst, nitrogen monoxide, NO, forms two other oxides of nitrogen, dinitrogen monoxide, N2O\mathrm{N}_{2} \mathrm{O}, and dinitrogen trioxide, N2O3\mathrm{N}_{2} \mathrm{O}_{3}.

…….NO( g)→……N2O( g)+……N2O3( g)ΔH⊖=−195.2 kJ mol−1ΔG⊖=−102.8 kJ mol−1\ldots \ldots . \mathrm{NO}(\mathrm{~g}) \rightarrow \ldots \ldots \mathrm{N}_{2} \mathrm{O}(\mathrm{~g})+\ldots \ldots \mathrm{N}_{2} \mathrm{O}_{3}(\mathrm{~g}) \quad \begin{aligned} \Delta H^{\ominus} & =-195.2 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \Delta G^{\ominus} & =-102.8 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}
[ 3 ]
(i)

Calculate the entropy change for the reaction at 298 K . Include the units in your answer.

ΔS⊖= units =\begin{aligned} \Delta S^{\ominus} & = \\ \text { units } & = \end{aligned}
[ 2 ]
(ii)

State whether the sign of ΔS⊖\Delta S^{\ominus} calculated in (iii) agrees with that predicted from your balanced equation in (i). Explain your answer.

[ 1 ]
(a)

Zinc metal can be obtained in a two-step process as shown.

step 12ZnS(s)+3O2( g)→2ZnO(s)+2SO2( g)12 \mathrm{ZnS}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{ZnO}(\mathrm{s})+2 \mathrm{SO}_{2}(\mathrm{~g})

step 2ZnO(s)+C(s)→Zn(l)+CO(g)2 \mathrm{ZnO}(\mathrm{s})+\mathrm{C}(\mathrm{s}) \rightarrow \mathrm{Zn}(\mathrm{l})+\mathrm{CO}(\mathrm{g})

The reactions are carried out at 800∘C800^{\circ} \mathrm{C}.

[ 1 ]
(i)

Predict the sign of the entropy change, ΔS⊖\Delta S^{\ominus}, of the reaction in step 1.

Explain your answer.

[ 1 ]
(ii)

Use the data in Table 1.1 to calculate ΔS⊖\Delta S^{\ominus} of the reaction shown in step 2.

Table 1.1

Table 1.1

JK−1 mol−1\mathrm{JK}^{-1} \mathrm{~mol}^{-1}
[Maximum number: 2]

Potassium chloride, KCl , and magnesium chloride, MgCl2\mathrm{MgCl}_{2}, are both ionic solids.

Table 1.1

Table 1.1

(a)
(i)

Explain what is meant by entropy, S.

[ 1 ]
(ii)

Potassium chloride is very soluble in water at 20∘C20^{\circ} \mathrm{C}.

Explain the solubility of potassium chloride by reference to change in entropy, ΔS\Delta S.

[ 1 ]
[Maximum number: 1]

Calcium chloride, CaCl2\mathrm{CaCl}_{2}, is an ionic solid.
The values of some energy changes are shown in Table 1.1.

Table 1.1

Table 1.1

(a)

Calcium fluoride, CaF2( s)\mathrm{CaF}_{2}(\mathrm{~s}), can be synthesised directly from its elements.

The value of ΔHf⊖(CaF2( s))\Delta H_{\mathrm{f}}^{\ominus}\left(\mathrm{CaF}_{2}(\mathrm{~s})\right) is −1214 kJ mol−1-1214 \mathrm{~kJ} \mathrm{~mol}^{-1}.

[ 1 ]
(i)

Predict the sign of the entropy change, ΔS⊖\Delta S^{\ominus}, for this synthesis. Explain your answer.

The sign of the entropy change is .
explanation

[ 1 ]
[Maximum number: 2]

Potassium chloride, KCl , and magnesium chloride, MgCl2\mathrm{MgCl}_{2}, are both ionic solids.

Table 1.1

Table 1.1

(a)
(i)

Explain what is meant by entropy, S.

[ 1 ]
(ii)

Potassium chloride is very soluble in water at 20∘C20^{\circ} \mathrm{C}.

Explain the solubility of potassium chloride by reference to change in entropy, ΔS\Delta S.

[ 1 ]
(a)

The standard entropy change for reaction 1 is ΔSr1⊖\Delta S_{\mathrm{r} 1}^{\ominus}.

Suggest, with a reason, how the standard entropy change for reaction 3 might compare with ΔSr1e\Delta S_{r 1}^{e}.

[ 2 ]
[Maximum number: 4]

Silicon tetrachloride, SiCl4\mathrm{SiCl}_{4}, is formed when silicon reacts with chlorine under suitable conditions. It is a colourless liquid with a low boiling point.

(a)
(i)

Silicon tetrachloride can be prepared according to reaction 1.
reaction 1

Si( s)+2Cl2( g)→SiCl4(I)ΔS⊖=−225.7JK−1 mol−1\mathrm{Si}(\mathrm{~s})+2 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{SiCl}_{4}(\mathrm{I}) \quad \Delta S^{\ominus}=-225.7 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}
Table

Calculate the standard entropy of chlorine, S⊖Cl2( g)S^{\ominus} \mathrm{Cl}_{2}(\mathrm{~g}). Show all your working.

S⊖Cl2( g)=\mathrm{S}^{\ominus} \mathrm{Cl}_{2}(\mathrm{~g})=
JK−1 mol−1\mathrm{JK}^{-1} \mathrm{~mol}^{-1}
[ 3 ]
(ii)

Explain why the entropy change for reaction 1 is negative.

[ 1 ]
[Maximum number: 1]

Potassium iodide, KI, is used as a reagent in both inorganic and organic chemistry.

(a)

KI slowly oxidises in air, forming I2\mathrm{I}_{2}.

reaction 14KI(s)+2CO2( g)+O2( g)→2 K2CO3( s)+2I2( s)ΔH⊖=−203.4 kJ mol−11 \quad 4 \mathrm{KI}(\mathrm{s})+2 \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{~K}_{2} \mathrm{CO}_{3}(\mathrm{~s})+2 \mathrm{I}_{2}(\mathrm{~s}) \quad \Delta H^{\ominus}=-203.4 \mathrm{~kJ} \mathrm{~mol}^{-1}

Table 1.2 shows some data relevant to this question.

Table 1.2

Table 1.2

[ 1 ]
(i)

Calculate the standard entropy change, ΔS⊖\Delta S^{\ominus}, of reaction 1 .

ΔS⊖=…………………………………..JK−1 mol−1\Delta S^{\ominus}=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \mathrm{JK}^{-1} \mathrm{~mol}^{-1}
[ 1 ]
(a)

Entropy is a measure of the disorder of a system.

Describe and explain what happens to the entropy of a gas when the temperature is increased.

[ 2 ]
(b)

The table shows four reactions.

[ 3 ]
(i)

For each reaction, predict the sign of the entropy change, ΔS⊖\Delta S^{\ominus}. If you predict no entropy change, write 'no change' in the table below. The first one has been done for you.

Table
[ 2 ]
(ii)

Explain why the entropy change for the first process is negative.

[ 1 ]
(c)

Calculate the standard entropy change, ΔS⊖\Delta S^{\ominus}, for this reaction.

N2( g)+3H2( g)→2NH3( g)\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})

Standard entropies, S⊖S^{\ominus}, in JK−1 mol−1\mathrm{JK}^{-1} \mathrm{~mol}^{-1} are given.

Table
ΔS⊖………….................JK−1 mol−1\Delta S^{\ominus} \ldots \ldots \ldots \ldots . . . . . . . . . . . . . . . . . \mathrm{JK}^{-1} \mathrm{~mol}^{-1}
[ 3 ]
[Maximum number: 3]

One method of producing hydrogen from natural gas is the reaction between hydrogen sulfide and methane.

2H2 S( g)+CH4( g)⇌CS2( g)+4H2( g)2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})+\mathrm{CH}_{4}(\mathrm{~g}) \rightleftharpoons \mathrm{CS}_{2}(\mathrm{~g})+4 \mathrm{H}_{2}(\mathrm{~g})
(a)
(i)

Predict the sign of ΔS⊖\Delta S^{\ominus} for this reaction. Explain your answer.

2H2 S( g)+CH4( g)⇌CS2( g)+4H2( g)ΔH⊖=+241 kJ mol−12 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})+\mathrm{CH}_{4}(\mathrm{~g}) \rightleftharpoons \mathrm{CS}_{2}(\mathrm{~g})+4 \mathrm{H}_{2}(\mathrm{~g}) \quad \Delta \mathrm{H}^{\ominus}=+241 \mathrm{~kJ} \mathrm{~mol}^{-1}

The free energy change, ΔGe\Delta G^{e}, for this reaction at 1000 K is +51 kJ mol−1+51 \mathrm{~kJ} \mathrm{~mol}^{-1}.

[ 1 ]
(ii)

Calculate the value of ΔS⊖\Delta S^{\ominus} for this reaction, stating its units.

ΔS⊖=\Delta S^{\ominus}=

units

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