Question

The value of $K_c$ for the reaction :
$\ce{A + 3B<=>2C}$ at $400^°C$ is $0.5.$ Calculate the value of $K_P$

• A. $1.64 × 10^{-4}$
• B. $1.64 × 10^{-6}$
• C. $1.64 × 10^{-5}$
• D. $1.64 × 10^{-3}$

Answer 1

Answer: (A)

$K_{p}=K_{c}\left[RT\right]^{\Delta n}_{g}\,\Delta n=2-4=-2$
$T=673K, K_{c}=0.5,$
$R=0.082\,L.\,atm.\,mol^{-1}\,K^{-1}$
$K_{p}=0.5\times\left(0.082\times673\right)^{-2}
=1.64\times10^{-4}\,atm.$