Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is K_p for the given equilibrium ? 2HI (g) H_2 (g) + I_2 (g)

7.11     A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium?

                2HI_{(g)}\rightleftharpoons H_{2}_{(g)}+I_{2}_{(g)}

Answers (1)

best_answer

The initial pressure of HI is 0.2 atm . At equilibrium, it has a partial pressure of 0.04 atm. Therefore, a decrease in the pressure of HI  is 0.2 - 0.04 = 0.16.

The given reaction is:

2HI_{(g)}\rightleftharpoons H_{2}_{(g)}+I_{2}_{(g)}

HI_{(g)}\rightleftharpoons 1/2H_{2}_{(g)}+1/2I_{2}_{(g)}

At equilibrium,
\\pHI = 0.04\\ pH_2 = \frac{0.16}{2}=0.08\\ pI_2 = \frac{0.16}{2}=0.08

Therefore, 
K_p = \frac{pH_2\times pI_2}{p^2HI}
         \\=\frac{0.08 \times 0.08}{(0.04)^2}\\ =4

Posted by

manish

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE wants international boards reined in; letter to education ministry seeks directions for AIU
anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens

Latest Question