Question

The relation between $K_p$ and $K_c$ of a reversible reaction at constant temperature is $K_p=...............$

• A. $K_p=K_c(RT{)}^{\Delta n}$
• B. $K_p=K_c(RT{)}^{-\Delta n}$
• C. $K_p=K_c\left(RT\right)$
• D. $K_p=K_c$

Answer 1

Answer: (A)

$K_p=K_c(RT{)}^{\Delta n}$

The equilibrium constant based on partial pressures is $K_p=\frac{P_C^c{P}_D^d}{P_A^a{P}_B^b}$

$\Rightarrow P_A=n_{A}RT/V\Rightarrow P_A=\left[A\right]RT$

Substituting into the expression for $K_p=\frac{\left(\right[C\left]RT{)}^c\right(\left[D\right]RT{)}^d}{\left(\right[A\left]RT{)}^a\right(\left[B\right]RT{)}^b}$

The equilibrium constant based in terms of concentration is $K_c=\frac{\left[C{]}^c\right[D{]}^d}{\left[A{]}^a\right[B{]}^b}$

Therefore, $K_p=K_c\times (RT{)}^{(c+d+...)–(a+b+...)}$

The exponent in $RT$ is the sum of the stoichiometric coefficients for the reactants subtracted from the sum of the stoichiometric coefficients for the products, defined as $\Delta n$.

$K_p=K_c(RT{)}^{\Delta n}$