First slide

Relation between Kc, and Kp

Question

Consider the reversible reaction, $A (g) ⇌ B (g) + C (g); K_{c} = 1 M$ at 300K. Value of $K_{p}$ at the same temperature is

Easy

A

$246 atm$
B

$24.6 atm$
C

$246 {atm}^{2}$
D

$24.6 {atm}^{2}$

Solution

$K_{p} = K_{c} {(RT)}^{∆ n_{g}}; ∆ n_{g} = n_{p} (g) - n_{r} (g) = 2 - 1 = 1; K_{p} = 1 mole {lit}^{- 1} (0.0821 lit atm {mol}^{- 1} K^{- 1} \times 300 K); K_{p} = 24.63 atm$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App