2- H - 2 - O - -g- -2- Cl - 2-g- -longrightarrow 4H- Cl - -g- - O - 2-g- -The relationship between -K-p- and -K-c- the equilibrium is- K - p - K - c - K - p - K - c - -RT- - 2 - K - p - K - c - -RT- - 1 - K - p - K - c - -RT- - -2 -

Question

Toppr Admin · Accepted Answer

The relationship between Kp and Kc for any reaction is given by the relation-Kp-Kc-RT-x394-ng where -x394-ng is the change in the gaseous moles of the reaction-For the given reaction -x394-ng-4-1-x2212-2-2-1Therefore-xA0-Kp-Kc-RT-1Hence- option C is correct-

2H2O(g)+2Cl2(g)⟶4HCl(g)+O2(g)The relationship between Kp and Kc for the equilibrium is:Kp=KcKp=Kc(RT)2Kp=Kc(RT)1Kp=Kc(RT)−2

The relationship between Kp and Kc for any reaction is given by the relation:Kp=Kc(RT)Δng where Δng is the change in the gaseous moles of the reaction.For the given reaction Δng=(4+1)−(2+2)=1Therefore, Kp=Kc(RT)1Hence, option C is correct.

$2 H_{2} O_{(g)} + 2 {C l}_{2 (g)} ⟶ 4 H {C l}_{(g)} + O_{2 (g)}$

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

$K_{p} = K_{c}$
$K_{p} = K_{c} {(R T)}^{2}$
$K_{p} = K_{c} {(R T)}^{1}$
$K_{p} = K_{c} {(R T)}^{- 2}$