For a reversible gas-phase reaction, X(g) + Y(g) leftharpoons Z(g)...

The equilibrium constant in terms of concentration, $K_{c}$, is given by the ratio of the forward and backward rate constants:

$K_{c} = \dfrac{k_{f}}{k_{b}}$

Given that $k_{f} = 82 k_{b}$, we have:

$K_{c} = 82$

For the reaction $\mathrm{X(g)} + \mathrm{Y(g)} \rightleftharpoons \mathrm{Z(g)}$, the change in the number of moles of gas is:

$\Delta n_{g} = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 1 - (1 + 1) = -1$

The relationship between $K_{p}$ and $K_{c}$ is:

$K_{p} = K_{c} (RT)^{\Delta n_{g}}$

Substituting the given values ($R = 0.082 \mathrm{~L~atm~mol^{-1}~K^{-1}}$ and $T = 500 \mathrm{~K}$):

$K_{p} = 82 \times (0.082 \times 500)^{-1}$

$K_{p} = \dfrac{82}{41} = 2.0$

Answer: $2.0$