229177
For the reaction $2 A+B \rightleftharpoons \quad C, \Delta H=x$ cal, which one of the following conditions-would favour the yield of $\mathrm{C}$ on the basis of Le-Chatelier principle?
1 High pressure, high temperature
2 Only low temperature
3 High pressure, low temperature
4 Only low pressure
Explanation:
The reversible reaction is : $2 \mathrm{~A}+\mathrm{B} \rightleftharpoons \quad \mathrm{C} \quad \Delta \mathrm{H}=+\mathrm{x} \text { cal }$ When pressure is increased in an equilibrium reaction, then the reaction moves towards the less number of moles. So, after increasing the temperature, the reaction move in forward direction. As the reaction is endothermic, high temperature is favorable and low temperature is unfavourable.
VITEEE- 2011
Chemical Equilibrium
229179
Which of the following comments about the following equilibrium reaction is the most correct? $\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{NH}_{3}(\mathrm{aq}) \rightleftharpoons \mathrm{HSO}_{4}^{-}(\mathrm{aq})+\mathrm{NH}_{4}^{+}(\mathrm{aq})$
1 The equilibrium lies mostly to the right
2 The equilibrium lies mostly to the left
3 The equilibrium is exactly in between where the concentration of each of the reactants is exactly equal to the concentration of each of the products
4 The equilibrium depends on the temperature of the equilibrium raction
Explanation:
$\mathrm{H}_{2} \mathrm{SO}_{4}$ when dissolves in water dissociate almost completely and produces $\mathrm{H}^{+}$ions. These ions are taken by $\mathrm{NH}_{3}$ and its conjugate acid $\mathrm{NH}_{4}^{+}$forms of $\mathrm{NH}_{3}$. Thus, the equilibrium lies mostly to right side or forward reaction. $\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{NH}_{3}(\mathrm{aq}) \rightleftharpoons \mathrm{HSO}_{4}^{-}(\mathrm{aq})+\mathrm{NH}_{4}^{+}(\mathrm{aq})$
UPTU/UPSEE-2011
Chemical Equilibrium
229182
Which of the following is not a physical equilibrium?
In physical equilibrium there should be a change in the physical state of the compounds involved in the equilibrium. In $3 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \quad 2 \mathrm{O}_{3}(\mathrm{~g})$ there is no change in the physical state, $\mathrm{O}_{2}$ is gas and $\mathrm{O}_{3}$ is also in gaseous state. $\therefore$ It is not a physical equilibrium.
UPTU/UPSEE-2009
Chemical Equilibrium
229183
In which one of the following equilibria, the increase of pressure over the equilibrium will favour the backward reaction?
1 Decomposition equilibrium of $\mathrm{HI}$
2 Formation equilibrium of $\mathrm{SO}_{3}$
3 Decomposition equilibrium of $\mathrm{NH}_{3}$
4 Formation equilibrium of $\mathrm{PCl}_{5}$
Explanation:
In the decomposition reaction of $\mathrm{NH}_{3}$ the number of moles of products formed is greater that of reactant involved. $2 \mathrm{NH}_{3} \rightleftharpoons \mathrm{N}_{2}+3 \mathrm{H}_{2}$ According to Le-Chatelier's principal, the increase of pressure on a chemical equilibrium shift in that direction in which the number of gaseous molecules decreases and vice-versa thus in this reaction, increase in pressure favours the backward reaction.
229177
For the reaction $2 A+B \rightleftharpoons \quad C, \Delta H=x$ cal, which one of the following conditions-would favour the yield of $\mathrm{C}$ on the basis of Le-Chatelier principle?
1 High pressure, high temperature
2 Only low temperature
3 High pressure, low temperature
4 Only low pressure
Explanation:
The reversible reaction is : $2 \mathrm{~A}+\mathrm{B} \rightleftharpoons \quad \mathrm{C} \quad \Delta \mathrm{H}=+\mathrm{x} \text { cal }$ When pressure is increased in an equilibrium reaction, then the reaction moves towards the less number of moles. So, after increasing the temperature, the reaction move in forward direction. As the reaction is endothermic, high temperature is favorable and low temperature is unfavourable.
VITEEE- 2011
Chemical Equilibrium
229179
Which of the following comments about the following equilibrium reaction is the most correct? $\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{NH}_{3}(\mathrm{aq}) \rightleftharpoons \mathrm{HSO}_{4}^{-}(\mathrm{aq})+\mathrm{NH}_{4}^{+}(\mathrm{aq})$
1 The equilibrium lies mostly to the right
2 The equilibrium lies mostly to the left
3 The equilibrium is exactly in between where the concentration of each of the reactants is exactly equal to the concentration of each of the products
4 The equilibrium depends on the temperature of the equilibrium raction
Explanation:
$\mathrm{H}_{2} \mathrm{SO}_{4}$ when dissolves in water dissociate almost completely and produces $\mathrm{H}^{+}$ions. These ions are taken by $\mathrm{NH}_{3}$ and its conjugate acid $\mathrm{NH}_{4}^{+}$forms of $\mathrm{NH}_{3}$. Thus, the equilibrium lies mostly to right side or forward reaction. $\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{NH}_{3}(\mathrm{aq}) \rightleftharpoons \mathrm{HSO}_{4}^{-}(\mathrm{aq})+\mathrm{NH}_{4}^{+}(\mathrm{aq})$
UPTU/UPSEE-2011
Chemical Equilibrium
229182
Which of the following is not a physical equilibrium?
In physical equilibrium there should be a change in the physical state of the compounds involved in the equilibrium. In $3 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \quad 2 \mathrm{O}_{3}(\mathrm{~g})$ there is no change in the physical state, $\mathrm{O}_{2}$ is gas and $\mathrm{O}_{3}$ is also in gaseous state. $\therefore$ It is not a physical equilibrium.
UPTU/UPSEE-2009
Chemical Equilibrium
229183
In which one of the following equilibria, the increase of pressure over the equilibrium will favour the backward reaction?
1 Decomposition equilibrium of $\mathrm{HI}$
2 Formation equilibrium of $\mathrm{SO}_{3}$
3 Decomposition equilibrium of $\mathrm{NH}_{3}$
4 Formation equilibrium of $\mathrm{PCl}_{5}$
Explanation:
In the decomposition reaction of $\mathrm{NH}_{3}$ the number of moles of products formed is greater that of reactant involved. $2 \mathrm{NH}_{3} \rightleftharpoons \mathrm{N}_{2}+3 \mathrm{H}_{2}$ According to Le-Chatelier's principal, the increase of pressure on a chemical equilibrium shift in that direction in which the number of gaseous molecules decreases and vice-versa thus in this reaction, increase in pressure favours the backward reaction.
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Chemical Equilibrium
229177
For the reaction $2 A+B \rightleftharpoons \quad C, \Delta H=x$ cal, which one of the following conditions-would favour the yield of $\mathrm{C}$ on the basis of Le-Chatelier principle?
1 High pressure, high temperature
2 Only low temperature
3 High pressure, low temperature
4 Only low pressure
Explanation:
The reversible reaction is : $2 \mathrm{~A}+\mathrm{B} \rightleftharpoons \quad \mathrm{C} \quad \Delta \mathrm{H}=+\mathrm{x} \text { cal }$ When pressure is increased in an equilibrium reaction, then the reaction moves towards the less number of moles. So, after increasing the temperature, the reaction move in forward direction. As the reaction is endothermic, high temperature is favorable and low temperature is unfavourable.
VITEEE- 2011
Chemical Equilibrium
229179
Which of the following comments about the following equilibrium reaction is the most correct? $\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{NH}_{3}(\mathrm{aq}) \rightleftharpoons \mathrm{HSO}_{4}^{-}(\mathrm{aq})+\mathrm{NH}_{4}^{+}(\mathrm{aq})$
1 The equilibrium lies mostly to the right
2 The equilibrium lies mostly to the left
3 The equilibrium is exactly in between where the concentration of each of the reactants is exactly equal to the concentration of each of the products
4 The equilibrium depends on the temperature of the equilibrium raction
Explanation:
$\mathrm{H}_{2} \mathrm{SO}_{4}$ when dissolves in water dissociate almost completely and produces $\mathrm{H}^{+}$ions. These ions are taken by $\mathrm{NH}_{3}$ and its conjugate acid $\mathrm{NH}_{4}^{+}$forms of $\mathrm{NH}_{3}$. Thus, the equilibrium lies mostly to right side or forward reaction. $\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{NH}_{3}(\mathrm{aq}) \rightleftharpoons \mathrm{HSO}_{4}^{-}(\mathrm{aq})+\mathrm{NH}_{4}^{+}(\mathrm{aq})$
UPTU/UPSEE-2011
Chemical Equilibrium
229182
Which of the following is not a physical equilibrium?
In physical equilibrium there should be a change in the physical state of the compounds involved in the equilibrium. In $3 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \quad 2 \mathrm{O}_{3}(\mathrm{~g})$ there is no change in the physical state, $\mathrm{O}_{2}$ is gas and $\mathrm{O}_{3}$ is also in gaseous state. $\therefore$ It is not a physical equilibrium.
UPTU/UPSEE-2009
Chemical Equilibrium
229183
In which one of the following equilibria, the increase of pressure over the equilibrium will favour the backward reaction?
1 Decomposition equilibrium of $\mathrm{HI}$
2 Formation equilibrium of $\mathrm{SO}_{3}$
3 Decomposition equilibrium of $\mathrm{NH}_{3}$
4 Formation equilibrium of $\mathrm{PCl}_{5}$
Explanation:
In the decomposition reaction of $\mathrm{NH}_{3}$ the number of moles of products formed is greater that of reactant involved. $2 \mathrm{NH}_{3} \rightleftharpoons \mathrm{N}_{2}+3 \mathrm{H}_{2}$ According to Le-Chatelier's principal, the increase of pressure on a chemical equilibrium shift in that direction in which the number of gaseous molecules decreases and vice-versa thus in this reaction, increase in pressure favours the backward reaction.
229177
For the reaction $2 A+B \rightleftharpoons \quad C, \Delta H=x$ cal, which one of the following conditions-would favour the yield of $\mathrm{C}$ on the basis of Le-Chatelier principle?
1 High pressure, high temperature
2 Only low temperature
3 High pressure, low temperature
4 Only low pressure
Explanation:
The reversible reaction is : $2 \mathrm{~A}+\mathrm{B} \rightleftharpoons \quad \mathrm{C} \quad \Delta \mathrm{H}=+\mathrm{x} \text { cal }$ When pressure is increased in an equilibrium reaction, then the reaction moves towards the less number of moles. So, after increasing the temperature, the reaction move in forward direction. As the reaction is endothermic, high temperature is favorable and low temperature is unfavourable.
VITEEE- 2011
Chemical Equilibrium
229179
Which of the following comments about the following equilibrium reaction is the most correct? $\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{NH}_{3}(\mathrm{aq}) \rightleftharpoons \mathrm{HSO}_{4}^{-}(\mathrm{aq})+\mathrm{NH}_{4}^{+}(\mathrm{aq})$
1 The equilibrium lies mostly to the right
2 The equilibrium lies mostly to the left
3 The equilibrium is exactly in between where the concentration of each of the reactants is exactly equal to the concentration of each of the products
4 The equilibrium depends on the temperature of the equilibrium raction
Explanation:
$\mathrm{H}_{2} \mathrm{SO}_{4}$ when dissolves in water dissociate almost completely and produces $\mathrm{H}^{+}$ions. These ions are taken by $\mathrm{NH}_{3}$ and its conjugate acid $\mathrm{NH}_{4}^{+}$forms of $\mathrm{NH}_{3}$. Thus, the equilibrium lies mostly to right side or forward reaction. $\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{NH}_{3}(\mathrm{aq}) \rightleftharpoons \mathrm{HSO}_{4}^{-}(\mathrm{aq})+\mathrm{NH}_{4}^{+}(\mathrm{aq})$
UPTU/UPSEE-2011
Chemical Equilibrium
229182
Which of the following is not a physical equilibrium?
In physical equilibrium there should be a change in the physical state of the compounds involved in the equilibrium. In $3 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \quad 2 \mathrm{O}_{3}(\mathrm{~g})$ there is no change in the physical state, $\mathrm{O}_{2}$ is gas and $\mathrm{O}_{3}$ is also in gaseous state. $\therefore$ It is not a physical equilibrium.
UPTU/UPSEE-2009
Chemical Equilibrium
229183
In which one of the following equilibria, the increase of pressure over the equilibrium will favour the backward reaction?
1 Decomposition equilibrium of $\mathrm{HI}$
2 Formation equilibrium of $\mathrm{SO}_{3}$
3 Decomposition equilibrium of $\mathrm{NH}_{3}$
4 Formation equilibrium of $\mathrm{PCl}_{5}$
Explanation:
In the decomposition reaction of $\mathrm{NH}_{3}$ the number of moles of products formed is greater that of reactant involved. $2 \mathrm{NH}_{3} \rightleftharpoons \mathrm{N}_{2}+3 \mathrm{H}_{2}$ According to Le-Chatelier's principal, the increase of pressure on a chemical equilibrium shift in that direction in which the number of gaseous molecules decreases and vice-versa thus in this reaction, increase in pressure favours the backward reaction.