- 1(current)
Q. No.At a temperature above 1073 K, coke can be used to reduce FeO to Fe. It can be best explained by:
1. Ellingham diagram.
2. Orgel diagram.
3. Tunabo-Sugano diagram.
4. Born-Haber cycle.
Consider the Ellingham diagram given below.
The Y-axis, and X-axis in the graph are respectively:
1. Gibbs energy(\(\Delta G_{r}^{o} \)), and Pressure
2. Gibbs energy (\(\Delta G_{r}^{o} \)), and temperature
3. Enthalpy(\(\Delta H_{r}^{o} \)), and temperature
4. Temperature, and Gibbs energy (\(\Delta H_{r}^{o} \))
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. |
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.
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. |
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
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Q. No.
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