Q. No.
Consider the following reactions at 1000 oC.
(I) ZnS + 1/2O2(g) \(\xrightarrow[]{\Delta }\) ZnO(g); ∆Go = -360 kJ mol-1
(II) Cs + 1/2O2(g) \(\xrightarrow[]{\Delta }\) CO(g); ∆Go = -460 kJ mol-1
Choose the correct statement about the above reactions at 1000 oC:
1. ZnO is more stable than CO
2. ZnO can be reduced to Zn by C
3. ZnO, and CO are formed at an equal rate
4. ZnO can not be reduced to Zn by C
The correct statement among the following is:
| 1. | In the decomposition of oxide into oxygen and gaseous metal, entropy increases. |
| 2. | Decomposition of oxide is an endothermic change. |
| 3. | To make ∆G° negative, the temperature should be high enough so that T∆S° > ∆H°. |
| 4. | All of the above. |
Assertion : The reduction of a metal oxide is easier if the metal formed is in liquid state
at the temperature of reduction.
Reason : The value of entropy change +S of the reduction process is more on +ve side when the metal formed is in liquid state and the metal oxide being reduced is in solid state. Thus the value of becomes more on negative side.
- If both the assertion and the reason are true and the reason is a correct explanation of the assertion
- If both the assertion and reason are true but the reason is not a correct explanation of the assertion
- If the assertion is true but the reason is false
- If both the assertion and reason are false
Considering Ellingham diagram, which of the following metals can be used to reduce alumina?
1. Fe
2. Zn
3. Mg
4. Cu
| 1. | Carbon and hydrogen act as suitable reducing agents for metal sulphides |
| 2. | The \(\Delta_f G^0\) of the sulphide is greater than those for \(C S_2\) and \(\mathrm{H}_2 \mathrm{~S}\) |
| 3. | The \(\Delta_f G^0\) is negative for roasting of sulphur ore to oxide. |
| 4. | Roasting of the sulphide to the oxide is thermodynamically feasible |
The correct statement among the following is:
| 1. | The rate of reaction cannot be understood from the Ellingham diagram. |
| 2. | During the formation of metal oxide \(\Delta S\) becomes negative and \(\Delta G\) becomes positive resulting in a positive slope. |
| 3. | There is an abrupt change in the slope of the Ellingham line when a change in phase (s→l) or (l→g) takes place. |
| 4. | All of the above. |
Ellingham's diagram for the formation of is a straight line in the given graph.
This is due to:
1. Increase in entropy during formation.
2. Decrease in entropy during formation.
3. Entropy remains constant during formation.
4. Can not be predicted.
A metal's oxide that can be reduced by Fe as a reducing agent at a temperature is:
1. Zr
2. Ca
3. Mg
4. None of the above.
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