(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

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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

The reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)} + q$ is favored by:

The equilibrium $Cr_2O_7^{2-} \rightleftharpoons 2CrO_4^{2-}$ shifts to the right in:

When $KOH$ is dissolved in water,heat is evolved. If the temperature is raised,the solubility of $KOH$

The dissociation of $NO$ increases with $.......$.

For the equilibrium reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$,the forward reaction will be favored if:

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