Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • One of the reaction that takes place in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and CO_2. FeO (s) + CO (g) Fe (s) + CO_2 (g); K_p = 0.265 atm at 1050K

7.20    One of the reaction that takes place in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and CO2.FeO_{(s)}+CO_{(g)}\rightleftharpoons Fe_{(s)}+CO_{2}_{(g)}; Kp = 0.265 atm at 1050K What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are: p_{CO}= 1.4 atm and pCO_2  = 0.80 atm?

Answers (1)

best_answer

We have,

K_p=0.265
the initial pressure of CO and CO_2 are 1.4 atm and 0.80atm resp.

The given reaction is-

                                FeO_{(s)}+CO_{(g)}\rightleftharpoons Fe_{(s)}+CO_{2}_{(g)}
initially,                                   1.4 atm                    0.80 atm

Q_p = \frac{pCO_2}{pCO}=\frac{0.80}{1.4} = 0.571
Since Q_p>K_c the reaction will proceed in the backward direction to attain equilibrium. The partial pressure of CO_2 will increase = decrease in the partial pressure of CO_2   = p                      

\therefore K_p = \frac{pCO_2}{pCO} = \frac{0.80-p}{1.4+p}=0.265
                                   =0.371+.265p=0.80-p

                                    By solving the above equation we get the value of p = 0.339 atm

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 board exams twice a year from 2026-27; 2 subject groups, 'better of two marks' in draft policy
anam-khan

CBSE Class 12 physics paper ‘extremely hard’, leaves students jittery; board gets court case warning
anam-khan

CBSE to felicitate scribes helping CWSN candidates in board exams; revises affiliation rules

Latest Question