Question #64924

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 degree C, a value of Kp = 1.91 has been established for this decomposition. If 0236 moles of N2O3 are placed in a 1.52-L vessel at 25 degree C, calculate the equilibrium partial pressures of N2O3(g), NO2(g) and NO(g).

Expert's answer

N2O3 = NO2 + NO

First, the mol of each component can be calculated from the equation [NO2][NO]/[N2O3] = Kp. Thus, C(N2O3) = 0.026 mol, C(NO2) = 0.210 mol and C(NO) = 0.210 mol.

According to a perfect gas state equation p = n*R*T/V. Thus, p(N2O3) = 42.4 kPa, p(NO2) = 342 kPa and p(NO) = 342 kPa.

First, the mol of each component can be calculated from the equation [NO2][NO]/[N2O3] = Kp. Thus, C(N2O3) = 0.026 mol, C(NO2) = 0.210 mol and C(NO) = 0.210 mol.

According to a perfect gas state equation p = n*R*T/V. Thus, p(N2O3) = 42.4 kPa, p(NO2) = 342 kPa and p(NO) = 342 kPa.

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