In a chemical equilibrium A + B rightleftharp | vedclass

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(B) The chemical equation is $A + B \rightleftharpoons C + D$. Initially,the moles are: $n(A) = 1, n(B) = 1, n(C) = 0, n(D) = 0$. At equilibrium,$0.5

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$1$

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(B) The chemical equation is $A + B \rightleftharpoons C + D$. Initially,the moles are: $n(A) = 1, n(B) = 1, n(C) = 0, n(D) = 0$. At equilibrium,$0.5 \, mol$ of each product is formed. So,the moles at equilibrium are: $n(C) = 0.5, n(D) = 0.5$. Since $1 \, mol$ of $A$ and $B$ produce $1 \, mol$ of $C$ and $D$,the moles of reactants consumed are $0.5 \, mol$ each. Thus,at equilibrium: $n(A) = 1 - 0.5 = 0.5$ and $n(B) = 1 - 0.5 = 0.5$. The equilibrium constant $K_c$ is given by the expression: $K_c = \frac{[C][D]}{[A][B]}$. Assuming the volume of the container is $V$,the concentrations are: $[C] = \frac{0.5}{V}, [D] = \frac{0.5}{V}, [A] = \frac{0.5}{V}, [B] = \frac{0.5}{V}$. Substituting these values: $K_c = \frac{(0.5/V)(0.5/V)}{(0.5/V)(0.5/V)} = 1$. Therefore,the equilibrium constant is $1$.

In a chemical equilibrium $A + B \rightleftharpoons C + D$,when $1 \, mol$ each of two reactants are mixed,$0.5 \, mol$ each of the products are formed. The equilibrium constant is

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