(A) With an increase in temperature from $273 \ K$ to $298 \ K$,the value of the equilibrium constant $(K)$ decreases from $2$ to $1 \times 10^{-2}$.A

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

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

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

A
exothermic
B
endothermic
C
unpredictable
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

$H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,$\Delta H = +q \ cal$. The formation of $HI$ is:

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On heating a mixture of $SO_2Cl_2$ and $CO$,two equilibria are simultaneously established: $SO_2Cl_{2(g)} \rightleftharpoons SO_{2(g)} + Cl_{2(g)}$ and $CO_{(g)} + Cl_{2(g)} \rightleftharpoons COCl_{2(g)}$. On adding more $SO_2$ at equilibrium,what will happen?

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$aA \rightleftharpoons mB + nC$,$\Delta H = + x \ kcal$. If $m + n < a$,then conditions favouring forward reaction will be

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The formation of $SO_3$ takes place according to the following reaction: $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$; $\Delta H = -45.2 \ kcal$. The formation of $SO_3$ is favoured by:

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In a vessel containing $SO_3$,$SO_2$,and $O_2$ at equilibrium,some helium gas is introduced so that the total pressure increases while temperature and volume remain constant. According to Le Chatelier's principle,the dissociation of $SO_3$:

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

Similar Questions

$H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,$\Delta H = +q \ cal$. The formation of $HI$ is:

On heating a mixture of $SO_2Cl_2$ and $CO$,two equilibria are simultaneously established: $SO_2Cl_{2(g)} \rightleftharpoons SO_{2(g)} + Cl_{2(g)}$ and $CO_{(g)} + Cl_{2(g)} \rightleftharpoons COCl_{2(g)}$. On adding more $SO_2$ at equilibrium,what will happen?

$aA \rightleftharpoons mB + nC$,$\Delta H = + x \ kcal$. If $m + n < a$,then conditions favouring forward reaction will be

The formation of $SO_3$ takes place according to the following reaction: $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$; $\Delta H = -45.2 \ kcal$. The formation of $SO_3$ is favoured by:

In a vessel containing $SO_3$,$SO_2$,and $O_2$ at equilibrium,some helium gas is introduced so that the total pressure increases while temperature and volume remain constant. According to Le Chatelier's principle,the dissociation of $SO_3$:

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