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.
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
The thermodynamic property useful for selecting the reducing agent is:
1. Gibbs free energy change .
2. Internal energy change.
3. Specific heat capacity.
4. None of the above.
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
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 (\(\Delta H_{r}^{o} \))
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
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. |
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
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.
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
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.
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
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. |
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
Al can reduce MgO under the following conditions:
1. Above 500 °C, Al can reduce MgO
2. Below 500 °C, Al can reduce MgO
3. Below 1350 °C, Al can reduce MgO
4. Above 1350 °C, Al can reduce MgO
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
Consider the following graph.
At 1400 oC, the \(\Delta G_{r}^{o}\) the value for the reaction
2FeO + 2C → 2Fe + CO is
1. -116 kJ mol-1
2. -106 kJ mol-1
3. 100 kJ mol-1
4. -100 kJ mol-1
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
© 2026 GoodEd Technologies Pvt. Ltd.