1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# At 1200∘C, the following equilibrium is established between chlorine atoms and molecule: Cl2(g)⇌2Cl(g)The composition of the equilibrium mixture may be determined by measuring the rate of effusion of the mixture through a pin hole. It is found that at 1200∘C and 1 atm pressure the mixture effuses 1.16 times as fast as krypton effuses under the same condition. The equilibrium constant Kc is:

A
6.71×104
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
6.92×104
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
7.71×104
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
7.34×104
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B 6.71×10−4The rate of effusion is inversely proportional to the molecular weight.The molecular weight of krypton is 83.8.1.161=√83.8MM=62.27Thus, the average molecular weight of the mixture of Cl2 and Cl is 62.27.Let x be the mole fraction of Cl2. The mole fraction of Cl will be 1−x.Hence, 71x+35.5(1−x)=62.27x=0.754The mole fractions of Cl2 and Cl are 0.754 and 0.246 respectively. They are also equal to the partial pressures as total pressure is 1 atm.Kp=PCl2PCl2=0.24620.754=0.08As Δn=1, KP=Kc(RT)Thus, 0.08=Kc(0.082×1473)Kc=6.71×10−4

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Equilibrium Constants
CHEMISTRY
Watch in App
Explore more
Join BYJU'S Learning Program