(A) The equilibrium constant $K$ decreases from $2$ to $1 \times 10^{-3}$ as the temperature increases from $273 \ K$ to $298 \ K$. According to Le Ch

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(A) The equilibrium constant $K$ decreases from $2$ to $1 \times 10^{-3}$ as the temperature increases from $273 \ K$ to $298 \ K$. According to Le Chatelier's principle,for an exothermic reaction,the equilibrium constant decreases with an increase in temperature. Since the value of $K$ decreases as the temperature increases,the forward reaction (formation of $C$ and $D$) is exothermic.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Le-Chaterlier principle and It’s application

Medium

The equilibrium constant for a reaction $A + B \rightleftharpoons C + D$ is $1 \times 10^{-3}$ at $298 \ K$ and is $2$ at $273 \ K$. The chemical reaction resulting in the formation of $C$ and $D$ is

A
exothermic
B
endothermic
C
can't be determined
D
there is no relationship between $\Delta H$ and $K$

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6-1.Equilibrium (Chemical Equilibrium)

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Le-Chaterlier principle and It’s application

Consider the reaction equilibrium,$2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$; $\Delta H^\circ = -198 \ kJ$. On the basis of Le Chatelier's principle,the condition favourable for the forward reaction is

EasyAIEEE 2003

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Assertion : $A$ catalyst does not alter the equilibrium constant of a reaction.
Reason : The catalyst forms a complex with the reactants and provides an alternate path with lower energy of activation for the reaction; the forward and the backward reactions are affected to the same extent.

MediumAIIMS 2010

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The equilibrium $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ shifts forward,if

Easy

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Explain the effect of adding $(i)$ Oxalic acid $(H_2C_2O_4)$,$(ii)$ $HgCl_2$,and $(iii)$ Potassium thiocyanate $(KSCN)$ on the equilibrium reaction: $Fe^{3+}(aq) + SCN^-(aq) \rightleftharpoons [Fe(SCN)]^{2+}(aq)$ (deep red color).

Difficult

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When $NaNO_3$ is heated in a closed vessel,$O_2$ is liberated and $NaNO_2$ is left behind. At equilibrium,

MediumIIT 1986

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The equilibrium constant for a reaction $A + B \rightleftharpoons C + D$ is $1 \times 10^{-3}$ at $298 \ K$ and is $2$ at $273 \ K$. The chemical reaction resulting in the formation of $C$ and $D$ is

Similar Questions

Consider the reaction equilibrium,$2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$; $\Delta H^\circ = -198 \ kJ$. On the basis of Le Chatelier's principle,the condition favourable for the forward reaction is

Assertion : $A$ catalyst does not alter the equilibrium constant of a reaction.Reason : The catalyst forms a complex with the reactants and provides an alternate path with lower energy of activation for the reaction; the forward and the backward reactions are affected to the same extent.

The equilibrium $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ shifts forward,if

Explain the effect of adding $(i)$ Oxalic acid $(H_2C_2O_4)$,$(ii)$ $HgCl_2$,and $(iii)$ Potassium thiocyanate $(KSCN)$ on the equilibrium reaction: $Fe^{3+}(aq) + SCN^-(aq) \rightleftharpoons [Fe(SCN)]^{2+}(aq)$ (deep red color).

When $NaNO_3$ is heated in a closed vessel,$O_2$ is liberated and $NaNO_2$ is left behind. At equilibrium,

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Assertion : $A$ catalyst does not alter the equilibrium constant of a reaction.
Reason : The catalyst forms a complex with the reactants and provides an alternate path with lower energy of activation for the reaction; the forward and the backward reactions are affected to the same extent.