The equilibrium constant for the reaction 2 HF (g) ⇒ H₂ (g) + F₂ (g) is 0.250 at a particular temperature. What is the equilibrium constant for the equation 12 H₂(g) + ½ F₂ (g) = HF (g)?

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9:10
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4
7
+/-
The equilibrium constant for the
reaction 2 HF (g) = H₂ (g) + F₂ (g)
is 0.250 at a particular
temperature. What is the
equilibrium constant for the
equation 12 H₂ (g) + ½ F₂ (g) = HF
(g)?
2
5
8
Question 1 of 10
.
3
60
9
O
Submit
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x 100
Transcribed Image Text:9:10 1 4 7 +/- The equilibrium constant for the reaction 2 HF (g) = H₂ (g) + F₂ (g) is 0.250 at a particular temperature. What is the equilibrium constant for the equation 12 H₂ (g) + ½ F₂ (g) = HF (g)? 2 5 8 Question 1 of 10 . 3 60 9 O Submit Tap here or pull up for additional resources XU x 100
Expert Solution
Step 1

Equilibrium constant for reverse reaction:

For reverse reaction the value of the equilibrium constant is the reciprocal of the equilibrium constant for forward reaction.

Equilibrium constant when coefficient are changed:

If a balanced chemical equation coefficient are multiplied by factor n, then  the original Equilibrium constant (K) is raised to the nth power to obtain the new equilibrium constant value.

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