Answer to Question #251472 in General Chemistry for Mallorie

Question

At 1261 degrees Celsius the equilibrium constant for the reaction: 2 IBr(g) —><— I2(g) + Br2(g)

is Kp= 1.09. If the initial pressure of IBr is 0.00985 atm, what are the equilibrium partial pressures of IBr, I2, and Br2?

1) p(IBr) = ___

2) p(I2) = ____

3) p(Br2) = ____

Answer 1

Moles of IBr= 2

Moles of I2 = 1

Volume = 67.2 L

Temperature = 273 + 1261 = 1534 K

PV = nRT

Partial pressure of IBR = 2×0.0821×208/67.2= 0.508atm

Partial pressure of I2= 1×0.0821×208/67.2= 0.254atm