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

# The equilibrium constants for the following reactions at 1400 K are given:2H2O(g)⇌2H2(g)+O2(g);K1=2.1×10−132CO2(g)⇌2CO(g)+O2(g);K2=1.4×10−12Then, the equilibrium constant K for the reaction, H2(g)+CO2(g)⇌CO(g)+H2O(g), is:

A
2.04
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
20.5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2.6
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
6.67
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is C 2.6(I) 2H2O(g)⇌2H2(g)+O2(g)∴K1=(pH2)2.pO2(pH2O)2=2.1×10−13→(I)(II) 2CO2(g)⇌2CO(g)+O2(g)∴K2=(pCO)2.pO2(pCO2)2=1.4×10−12→(II)Now, H2(g)+CO2(g)⇌CO(g)+H2O(g)K=pCO.pH2OpH2.pCO2=(K2)1/2×(1K1)1/2⇒K=(1.42.1×10−1210−13)1/2=(20/3)1/2=2.6

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