SME
General Chemistry II · Problem type
K of a reversed, scaled or added reaction
Use this for “K = 2.5 × 10⁻³ for A ⇌ 2B. What is K for 2B ⇌ A, and for ½A ⇌ B?”
This is an example. Type your own values to start a new problem.
K
=
K₁
×
K₂
Formulas
K
=
K₁
n
×
K₂
400
=
0.0025
-1
×
1
Assumptions
•
Reversing a reaction flips K over: n = −1 gives 1 ÷ K₁.
•
Multiplying every coefficient by n raises K to the power n: doubling squares it, halving takes its square root.
•
Adding two reactions multiplies their K values. With no second reaction, leave K₂ = 1.
•
Every K is at the same temperature.
Step-by-step
WE KNOW
K₁ = 0.0025, n = −1, K₂ = 1
FIND
equilibrium constant overall (K)
Step 1 · Find equilibrium constant overall (K)
Use K = K₁ⁿ × K₂
Raise
K₁
to
the
multiplier,
then
multiply
by
the
K
of
the
reaction
added.
K
=
0.0025
-1
×
1
K
=
400
Check
Put the numbers back in. Both sides should match.
400
=
0.0025
-1
×
1
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.