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At 1755 oC the equilibrium constant for the reaction: 2 IBr(g) I2(g) + Br2(g) is KP = 0.748. If the initial pressure of IBr is 0.00465 atm, what are the equilibrium partial pressures of IBr, I2, and Br2? p(IBr) = . p(I2) = . p(Br2) = .

Respuesta :

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From the calculation, the equilibrium partial pressure of IBr is 6.5 * 10^-4 atm while that of I2 and Br2 is 0.002 atm.

What is the equilibrium pressure?

We must set up the ICE table as shown hence;

              2 IBr(g) < ------> I2(g) + Br2(g)

I              0.00465            0        0

C                  - 2x                + x      + x

E            0.00465 - 2x       + x      + x

Kp = pI2. pBr2/pIBr^2

pI2 = pBr2 = x

0.748 = x^2/  0.00465 - 2x

0.748 (0.00465 - 2x) = x^2

3.5 * 10^-3 - 1.496x = x^2

x^2 + 1.496x  - 3.5 * 10^-3 = 0

x=0.002 atm

Hence;

For IBr =   0.00465 - 2(0.002) = 6.5 * 10^-4 atm

For I2 and Br2 = 0.002 atm

Learn more about  equilibrium constant:https://brainly.com/question/10038290

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