Question

If reaction C = D is at equilibrium with PC=0.5 atm and PD=1.0 atm, then PC is increased to 1.0 atm while PD remains the same. What is the new equilibrium partial pressure of D?

A. 0.67 atm

B. 1.00 atm

C. 1.33 atm

D. 1.67 atm

Answer 1

consider the reaction

C ---> D

Kp = ( pD / pC)

given

pD = 1 , pC = 0.5

so

Kp = 1 / 0.5

Kp = 2

now

C ---> D

now partial pressure of C is icreased to 1 atm

now

using ICE table

initial pressure of C , D are 1 , 1

change in pressure of C , D are -x , +x

equilibrium pressure of C , D are 1-x , 1 + x

now

Kp = (pD) / (pC)

2 = (1+x) / ( 1-x)

2 (1-x) = 1 + x

2 - 2x = 1 + x

1 = 3x

x = 1/3 = 0.333

now

pD = 1 + x = 1 + 0.33 = 1.33 atm

so

the new equilibrium partial pressure of D is 1.33 atm

so

the answer is C) 1.33 atm