Question

Liquid N2O3 is dark blue at low temperatures, but the color fades and becomes greenish at higher temperatures as the compound decomposes to NO and NO2. At 25 °C, a value of KP = 1.91 has been established for this decomposition. If 0.236 moles of N2O3 are placed in a 1.52-L vessel at 25 °C, calculate the equilibrium partial pressures of N2O3(g, NO2(g, and NO(g.

a) PN2O3 = 0.217atm , PNO2 = 0.479atm , PNO = 0.155atm
b) PN2O3 = 0.155atm , PNO2 = 0.479atm , PNO = 0.217atm
c) PN2O3 = 0.479atm , PNO2 = 0.217atm , PNO = 0.155atm
d) PN2O3 = 0.479atm , PNO2 = 0.155atm , PNO = 0.217atm

Answer 1

To find the equilibrium partial pressures of N2O3, NO2, and NO in the given reaction, set up an ICE table, calculate the change in moles for each substance, and use the Kp expression to find the equilibrium partial pressures.

Step-by-step explanation:

In this question, we are given the equilibrium constant, Kp, for the decomposition of N2O3 to NO and NO2. We are also given the initial amount of N2O3 and the volume of the vessel. To find the equilibrium partial pressures of N2O3, NO2, and NO, we need to set up an ICE table and use the Kp expression.

1. Set up an ICE table to find the change in moles for each substance.
2. Calculate the equilibrium partial pressure of each substance using the initial amount and the change in moles.
3. Plug the values into the Kp expression to find the equilibrium partial pressures of N2O3, NO2, and NO.

The correct answer is option d) PN2O3 = 0.479atm, PNO2 = 0.155atm, PNO = 0.217atm.