For the reaction I2 (g) + Br2 (g) <-------> 2IBr (g) Kc=280 at 150 degrees C. Su

Question

For the reaction I2 (g) + Br2 (g) <-------> 2IBr (g)
Kc=280 at 150 degrees C. Suppose that 0.460 mol IBr in 1.00 L flask is allowed to reach equilibrium at 150 degrees C.

What is the equilibrium for IBr?
What is the equilibrum for I2?
What is the equilibrium for Br2?

Explanation / Answer

Kc = [I2] * [Br2]/ [IBr]^2 I2 + Br2 IBr Initial Concentrations: 0 0 0.46 +x +x -x 280 = [x] * [x]/ [0.46 -x]^2 280 * ( 0.46^2 +2*.46*-x + x^2) = x^2 I2 + Br2 ---> 2IBr since in the original there's no I2 nor Br2, the equilibrium will move to the left side. Lets assume x moles of I2 and x moles of Br2 formed in the equilibrium state, so moles of 2IBr will become 0.46-2x. Then use equilibrium constant equation to find x, K = [IBr]^2 / [I2][Br2] 280 = (0.46-2x)^2 / x^2 16.73 = (0.46-2x)/x x = 0.0246 mole/L. So at equilibrium, [I2]=[Br2]=0.0246 [IBr]=0.46-2x=0.4109

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