Question

the equilibrium constant, ( k_{c} ), for the following reaction is ( 6.50\times10^{-3} ) at ( 298 k ). calculate ( k_{p} ) for this reaction at this temperature.

( 2nobr(g)
ightleftharpoons 2no(g)+br_{2}(g) )

( k_{p}= )

Explanation:

Step1: Determine the value of $\Delta n$

For the reaction \(2NOBr(g)
ightleftharpoons2NO(g)+Br_{2}(g)\), \(\Delta n=(2 + 1)-2=1\)

Step2: Use the formula \(K_{p}=K_{c}(RT)^{\Delta n}\)

Given \(K_{c}=6.50\times 10^{-3}\), \(R = 0.0821\space L\cdot atm/(mol\cdot K)\), \(T=298\space K\), and \(\Delta n = 1\)
Substitute the values into the formula:
\(K_{p}=6.50\times 10^{-3}\times(0.0821\times298)^{1}\)
First calculate \(0.0821\times298=24.4658\)
Then \(K_{p}=6.50\times 10^{-3}\times24.4658\)
\(K_{p}=0.159\)

Answer:

\(0.159\)