For the reaction- CO-g-Cl-2-g-rightleftharpoons COCl-2-g- the dfrac -K-p-K-c- is equal toRT1sqrt RTfrac -1-RT-

Question

Toppr Admin · Accepted Answer

The correct option is C 1RTKp-Kc-RT-x394-n-Here-xA0-x394-n- Moles of gaseous product -x2212- Moles of gaseous reactant-xA0-x394-n-1-x2212-2-x2212-1Kp - Kc-RT-x2212-1-x2234-KpKc-1RT

For the reaction, $C O (g) + C l_{2} (g) ⇌ C O C l_{2} (g)$ the $\frac{K_{p}}{K_{c}}$ is equal to
RT
1
$\sqrt{R} T$
$\frac{1}{R T}$

The correct option is C $\frac{1}{R T}$
$K_{p} = K_{c} (R T)^{Δ n}$ ;
Here,
$Δ n =$ Moles of gaseous product $-$ Moles of gaseous reactant

$Δ n = 1 - 2 = - 1$

$K_{p}$ = $K_{c}$ (RT) $^{- 1}$

$∴ \frac{K_{p}}{K_{c}} = \frac{1}{R T}$

For the reaction, CO(g)+Cl2(g)⇌COCl2(g) the KpKc is equal toRT1√RT1RT

The correct option is C 1RTKp=Kc(RT)Δn;Here, Δn= Moles of gaseous product − Moles of gaseous reactant Δn=1−2=−1Kp = Kc(RT)−1∴KpKc=1RT

For the reaction, $C O (g) + C l_{2} (g) ⇌ C O C l_{2} (g)$ the $\frac{K_{p}}{K_{c}}$ is equal to
RT
1
$\sqrt{R} T$
$\frac{1}{R T}$

The correct option is C $\frac{1}{R T}$
$K_{p} = K_{c} (R T)^{Δ n}$ ;
Here,
$Δ n =$ Moles of gaseous product $-$ Moles of gaseous reactant

$Δ n = 1 - 2 = - 1$

$K_{p}$ = $K_{c}$ (RT) $^{- 1}$

$∴ \frac{K_{p}}{K_{c}} = \frac{1}{R T}$