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For the reaction A (g) → 2 B (g), K = 14.7 at 298 K. What is the value of Q for this reaction at 298 K when ∆G = -20.5 kJ/mol?

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Answer:

Q= 250= 2,5 x 10^2

Explanation:

ΔG=ΔGº +RTlnQ. R=8,314 J/molK and T=298K

In the balance ΔG=ΔGº +RTlnK=0 → ΔGº= -RTLnK= - 6,7 KJ/mol

ΔG=ΔGº +RTlnQ → -20,5 =  -6,7  + 2,5 LnQ

→LnQ= -5,5 → Q=250

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