Question

For the reversible reaction,

$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$

at ${500}^{o}C$, the value of $K_p$ is $1.44\times {10}^{-5}$ If all the partial pressures measured in atm and volume in a liter, then find the value of $K_c$ .

• A. $\frac{1.44\times {10}^{-5}}{(0.082\times 500{)}^{-2}}$
• B. $\frac{1.44\times {10}^{-5}}{(8.314\times 773{)}^{-2}}$
• C. $\frac{1.44\times {10}^{-5}}{(0.082\times 773{)}^2}$
• D. $\frac{1.44\times {10}^{-5}}{(0.082\times 773{)}^{-2}}$

Answer 1

Answer: (D)

$\frac{1.44\times {10}^{-5}}{(0.082\times 773{)}^{-2}}$

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

Given : T = 500${}^{0}C$= 500+273 = 773 Kelvin

R = 0.082 $Latm$ $K^{-1}mol{e}^{-1}$

$\Delta n=2-4=-2$

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

$K_c=\frac{1.44\times {10}^{-}5}{(0.082\times 773{)}^{-2}}$