At 1425 oC the equilibrium constant for the reaction: 2 IBr(g) I2(g) + Br2(g) is KP...

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At 1425 oC the equilibrium constant for the reaction: 2 IBr(g) I2(g) + Br2(g) is KP...

At 1425 oC the equilibrium constant for the reaction: 2 IBr(g) I2(g) + Br2(g) is KP = 0.937.

If the initial pressure of IBr is 0.00957 atm, what are the equilibrium partial pressures of IBr, I2, and Br2?

p(IBr) = ____.

p(I2) = _____.

p(Br2) = _____ .

science chemistry

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Answer 1

Answer #1

2 I Bora I 0.00gb 7 . – 2x 150.00957-24 Iz & 0 tx Bra 0 +2 0.00957-2x x 2 ke x? (0.00 957 -2x)2 1 = (0.967)(0.00957–2x) 1.967

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Answer 2

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At 1425 oC the equilibrium constant for the reaction: 2 IBr(g) I2(g) + Br2(g) is KP...

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At 1425 oC the equilibrium constant for the reaction: 2 IBr(g) I2(g) + Br2(g) is KP...

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At 659 oC the equilibrium constant for the reaction: 2 IBr(g) I2(g) + Br2(g) is KP...

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At 1425 oC the equilibrium constant for the reaction: 2 IBr(g) I2(g) + Br2(g) is KP...

Homework Answers

Add Answer to: At 1425 oC the equilibrium constant for the reaction: 2 IBr(g) I2(g) + Br2(g) is KP...

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Add Answer to:
At 1425 oC the equilibrium constant for the reaction: 2 IBr(g) I2(g) + Br2(g) is KP...