Step 1: For the equilibrium reaction to have \( K_p = K_c \), the relationship between pressure and concentration should be the same. This happens only when the total number of moles of reactants and products on both sides of the equation are equal, i.e., no change in volume.
Step 2: Consider the reaction: \[ A_2(g) + B_2(g) \rightleftharpoons 2AB(g) \] In this case, the number of moles of reactants is \( 1 + 1 = 2 \), and the number of moles of products is also 2. This results in the equality of \( K_p \) and \( K_c \), as the mole ratio of reactants and products is equal.
Step 3: Thus, the correct reaction for which \( K_p = K_c \) is: \[ A_2(g) + B_2(g) \rightleftharpoons 2AB(g) \]
The equilibrium constant for decomposition of $ H_2O $ (g) $ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1}) $ is $ 8.0 \times 10^{-3} $ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ($ \alpha $) of water is _____ $\times 10^{-2}$ (nearest integer value). [Assume $ \alpha $ is negligible with respect to 1]
For the reaction A(g) $\rightleftharpoons$ 2B(g), the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K.
[Given: R = 0.0831 atm $mol^{–1} K^{–1}$]
$K_p$ for the reaction at 1000 K is:
Consider the following gas phase dissociation, PCl$_5$(g) $\rightleftharpoons$ PCl$_3$(g) + Cl$_2$(g) with equilibrium constant K$_p$ at a particular temperature and at pressure P. The degree of dissociation ($\alpha$) for PCl$_5$(g) is
PCl$_5$(g) $\rightleftharpoons$ PCl$_3$(g) + Cl$_2$(g)