For the equilibrium,$2 \ NOCl \ (g) \rightleftharpoons 2 \ NO \ (g) + Cl_{2} \ (g)$,the value of the equilibrium constant,$K_{c}$ is $3.75 \times 10^{-6}$ at $1069 \ K$. Calculate the $K_{p}$ for the reaction at this temperature?
(N/A) The relationship between $K_{p}$ and $K_{c}$ is given by the equation: $K_{p} = K_{c}(RT)^{\Delta n}$.
For the given reaction: $2 \ NOCl \ (g) \rightleftharpoons 2 \ NO \ (g) + Cl_{2} \ (g)$.
The change in the number of moles of gaseous products and reactants is $\Delta n = (2 + 1) - 2 = 1$.
Given $K_{c} = 3.75 \times 10^{-6}$,$R = 0.0831 \ \text{L bar K}^{-1} \text{mol}^{-1}$,and $T = 1069 \ K$.
Substituting the values: $K_{p} = 3.75 \times 10^{-6} \times (0.0831 \times 1069)^{1}$.
$K_{p} = 3.75 \times 10^{-6} \times 88.8339$.
$K_{p} \approx 3.33 \times 10^{-4}$.
For the given reaction: $2 \ NOCl \ (g) \rightleftharpoons 2 \ NO \ (g) + Cl_{2} \ (g)$.
The change in the number of moles of gaseous products and reactants is $\Delta n = (2 + 1) - 2 = 1$.
Given $K_{c} = 3.75 \times 10^{-6}$,$R = 0.0831 \ \text{L bar K}^{-1} \text{mol}^{-1}$,and $T = 1069 \ K$.
Substituting the values: $K_{p} = 3.75 \times 10^{-6} \times (0.0831 \times 1069)^{1}$.
$K_{p} = 3.75 \times 10^{-6} \times 88.8339$.
$K_{p} \approx 3.33 \times 10^{-4}$.
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