Question
In a reaction A+2B ⇋ 2C; 2 moles ofA, 3 mole of B and 2 mole of C are placed in a 2 L flask and the equilibrium concentration of C is 0.5 mole/L. The equilibrium constant K for the reaction is:
Answer: Option C
:
C
A+2B⇋2CInitialmoles232Molarconc.2/2=1mol/L3/2=1.5mol/L2/2=1mol/LAtEqui.Conc1+0.25=1.25mol/L1.5+0.5=2mol/L0.5mol/L
K=(0.5)2/(1.25×(2)2)=0.05 Note that concentration of C has decreased, i.e., to attain equilibrium, reaction has shifted backwards. Decrease in conc. of C will be 0.5mol/L and hence increase in concentration of B will be 0.5mol/L and that of A will be 0.25mol/L
Was this answer helpful ?
:
C
A+2B⇋2CInitialmoles232Molarconc.2/2=1mol/L3/2=1.5mol/L2/2=1mol/LAtEqui.Conc1+0.25=1.25mol/L1.5+0.5=2mol/L0.5mol/L
K=(0.5)2/(1.25×(2)2)=0.05 Note that concentration of C has decreased, i.e., to attain equilibrium, reaction has shifted backwards. Decrease in conc. of C will be 0.5mol/L and hence increase in concentration of B will be 0.5mol/L and that of A will be 0.25mol/L
Was this answer helpful ?
More Questions on This Topic :
Question 10. Which of the followings is incorrect? ....
Submit Solution