QUESTION 18 4 At 896 K, Kp = 0.471 for the equilibrium: 2 NOBr (g) = 2 NO (g) + Br2 (g) If the equilibrium pressures of NO and NOBr are equal, what is the equilibrium pressure of Bra

4. IBr decomposes according to the reaction at 150 C: 2 IBr (g) = I2 (g) + Br2 (g) If IBr is placed in a container at an initial pressure of 5.0 atm, what will be its equilibrium pressure at 150°C? K, = 8.3 x 10-3

The equilibrium constant for the following reaction is 67.6 at 232 °C. PCI 3(g) + Cl₂(g) PCI5(g) K = 67.6 at 232 °C Calculate the equilibrium constant for the following reactions at 232 °C. (a) PCI5(g) PCl3(g) + Cl₂(g) K= (b) 2 PCI5(g) 2 PCI3(g) + 2 Cl₂(g) K=

A student ran the following reaction in the laboratory at 588 K: CO(g) + Cl₂(g) —COCI₂(g) When she introduced CO(g) and Cl₂(g) into a 1.00 L evacuated container, so that the initial partial pressure of CO was 1.74 atm and the initial partial pressure of Cl2 was 0.951 atm, she found that the equilibrium partial pressure of COCI₂ was 0.685 atm. Calculate the equilibrium constant, Kp, she obtained for this reaction. Кр =

A student ran the following reaction in the laboratory at 591 K:COCl2(g) CO(g) + Cl2(g)When she introduced COCl2(g) at a pressure of 0.380 atm into a 1.00 L evacuated container, she found the equilibrium partial pressure of Cl2(g) to be 0.247 atm.Calculate the equilibrium constant, Kp, she obtained for this reaction.

Given the following reaction at equilibrium, if Kc= 1.90 x 10 19 at 25.0 °C, Kp =. H2 (g) +Br2 (g) 2HBr (g) 5.26 x 10-20 1.56 x 104 6.44 x 105 1.90 x 1019 none of these