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Equilibrium
Le Chatelier's Principle
5 questions from 5 papers, each with its answer.
- questions
- 5
- papers
- 5
- years asked, 2017–2026
- 2 of 10
- of Equilibrium
- 4.81%
Appears in
Concepts tested
- Effect of pressure and temperature 2
- K depends only on temperature
- Le Chatelier's Principle Application
- le-chatelier-temperature-pressure
All Questions
2025 1 question
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Higher yield of $\text{NO}$ in $\text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}(g)$ can be obtained at [$\Delta H = +180.7\text{ kJ mol}^{-1}$]:
A. higher temperature
B. lower temperature
C. higher concentration of $\text{N}_2$
D. higher concentration of $\text{O}_2$
Choose the correct answer from the options given below:The reaction is endothermic ($\Delta H > 0$), so higher temperature shifts equilibrium forward (A). Adding reactants (higher $[\text{N}_2]$ or $[\text{O}_2]$) also drives the forward reaction (C, D).
According to Le Chatelier's principle: 1. The reaction is endothermic in the forward direction ($\Delta H = +180.7\text{ kJ mol}^{-1}$). Raising the temperature favors the endothermic path, shifting equilibrium toward the product NO (A is true; B is false). 2. Increasing the concentration of reactants (either $\text{N}_2$ or $\text{O}_2$) drives the reaction forward to consume the added reactants and generate more NO (C and D are true). Therefore, higher yield is achieved under A, C, and D.
2018 1 question
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Which one of the following conditions will favour maximum formation of the product in the reaction, $A_2(g) + B_2(g) \rightleftharpoons X_2(g)$; $\Delta_rH = -X$ kJ?
Exothermic reaction with decrease in gas moles is favoured by low T and high P.
$A_2(g) + B_2(g) \rightleftharpoons X_2(g)$; $\Delta H = -X$ kJ On increasing pressure, equilibrium shifts in the direction where pressure decreases, i.e. the forward direction. On decreasing temperature, equilibrium shifts in the exothermic direction, i.e. the forward direction. So high pressure and low temperature favour maximum formation of product.
2014 AIPMT 1 question
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For the reversible reaction: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ + heat. The equilibrium shifts in forward direction:According to Le Chatelier's principle:
In exothermic reactions, low temperature favours the forward reaction.
On increasing pressure, the equilibrium shifts towards the side with fewer moles of gas.
So the forward reaction is favoured by increasing pressure and decreasing temperature.
2011 AIPMT-PRE 1 question
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The value of $\Delta H$ for the reaction $X_2(g) + 4Y_2(g) \rightleftharpoons 2XY_4(g)$ is less than zero. Formation of $XY_4(g)$ will be favoured at$\Delta H < 0$, so the forward reaction is exothermic. By Le Chatelier's principle, an exothermic reaction is favoured by low temperature.
Moles of gaseous reactants = 1 + 4 = 5; moles of gaseous products = 2.
$\Delta n_g = 2 - 5 = -3$, so the forward reaction takes place with a decrease in the number of gaseous moles (decrease in volume).
Increasing the pressure shifts the equilibrium towards the side with fewer gaseous moles, i.e., towards $XY_4$.
Hence formation of $XY_4$ is favoured at high pressure and low temperature.
2000 AIPMT 1 question
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For any reversible reaction, if concentration of reactants increases, then effect on equilibrium constant :Increasing the concentration of reactants shifts the equilibrium position forward (more products form), as predicted by Le Chatelier's principle.
However, the concentrations readjust so that the ratio $\dfrac{[\text{products}]}{[\text{reactants}]}$ returns to the same value. The equilibrium constant itself does not change.
The equilibrium constant changes only with temperature.
So the equilibrium constant remains unchanged.