Is 8.3 or 11.7 the Correct Equilibrium Constant?

Click For Summary
SUMMARY

The equilibrium constant (Kc) for the reaction 2A + B ⇌ C was calculated using the formula Kc = [C]/([A]^2[B]). Given the equilibrium concentrations of 0.2 moles of A, 0.45 moles of B, and 0.15 moles of C, the calculated Kc value is 8.3. However, there is a discrepancy with a provided answer of 11.7 in a multiple-choice question. The calculations confirm that 8.3 is correct based on the equilibrium concentrations provided, and the initial concentrations do not affect this outcome.

PREREQUISITES
  • Understanding of chemical equilibrium concepts
  • Familiarity with the equilibrium constant formula (Kc)
  • Basic knowledge of stoichiometry
  • Ability to perform calculations involving moles and concentrations
NEXT STEPS
  • Review the derivation and application of the equilibrium constant (Kc) in chemical reactions
  • Study the impact of initial concentrations on equilibrium states
  • Learn about common pitfalls in calculating Kc values
  • Explore advanced topics in chemical kinetics and equilibrium
USEFUL FOR

Chemistry students, educators, and professionals involved in chemical analysis and reaction dynamics will benefit from this discussion.

Aafia
Messages
70
Reaction score
1

Homework Statement


[/B]
The following reaction was allowed to reach the state of equilibrium
2A+B <====> C
The initial amounts of reactant present in one litre of solution were 0.5 mole of A and 0.6 mole of B. At equilibrium the amounts were 0.2 mole of A and 0.45 mole of B and 0.15 mole of C. Calculate equilibrium constant

Homework Equations



Kc= [product]/[reactants]

The Attempt at a Solution


[/B]
Kc= [0.15]/0.2^2×0.45
Kc=0.15/0.018
= 8.3
In my book one of the option of mcq given is 8.3 but the option marked as correct is 11.7 which one is right? If 11.7 is correct then how?
 
Physics news on Phys.org
Aafia said:

Homework Statement


[/B]
The following reaction was allowed to reach the state of equilibrium
2A+B <====> C
The initial amounts of reactant present in one litre of solution were 0.5 mole of A and 0.6 mole of B. At equilibrium the amounts were 0.2 mole of A and 0.45 mole of B and 0.15 mole of C. Calculate equilibrium constant

Homework Equations



Kc= [product]/[reactants]

The Attempt at a Solution


[/B]
Kc= [0.15]/0.2^2×0.45
Kc=0.15/0.018
= 8.3
In my book one of the option of mcq given is 8.3 but the option marked as correct is 11.7 which one is right? If 11.7 is correct then how?

I do not see any reason your answer be wrong.

The question contains the superfluous information of initial concentrations, but as you can check these are consistent with the final.
 

Similar threads

Replies
15
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 19 ·
Replies
19
Views
4K
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
5
Views
2K