Question

consider the general reversible reaction.
aa + bb \longleftrightarrow cc + dd
what is the equilibrium constant expression for the given system?
\bigcirc \\ k_{eq} = \frac{ccdd}{aabb}
\bigcirc \\ k_{eq} = \frac{cd}{ab}
\bigcirc \\ k_{eq} = \frac{a^ab^b}{c^cd^d}
\bigcirc \\ k_{eq} = \frac{c^cd^d}{a^ab^b}

Explanation:

Step1: Recall Equilibrium Constant Formula

For a reversible reaction \( aA + bB
ightleftharpoons cC + dD \), the equilibrium constant \( K_{eq} \) is defined as the ratio of the product of the molar concentrations of the products (each raised to the power of their stoichiometric coefficients) to the product of the molar concentrations of the reactants (each raised to the power of their stoichiometric coefficients).

Step2: Apply the Formula

The products are \( C \) (with coefficient \( c \)) and \( D \) (with coefficient \( d \)), so their concentrations are \([C]^c\) and \([D]^d\) respectively. The reactants are \( A \) (with coefficient \( a \)) and \( B \) (with coefficient \( b \)), so their concentrations are \([A]^a\) and \([B]^b\) respectively. Thus, \( K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b} \).

Answer:

\( K_{eq} = \frac{[C]^c[D]^d}{[A]^a[B]^b} \) (the last option among the given choices)