Be sure to answer all parts. The equilibrium constant (KP) for the formation of

Question

Be sure to answer all parts. The equilibrium constant (KP) for the formation of the air pollutant nitric oxide (NO) in an automobile engine at 537C is 4.5 × 10-11. N2(g) + O2(g) 2NO(g) (a) Calculate the partial pressure of NO under these conditions if the partial pressures of nitrogen and oxygen are 3.00 and 0.012 atm, respectively. × 10 atm (b) Repeat the calculation for atmospheric conditions in which the partial pressures of nitrogen and oxygen are 0.78 and 0.21 atm, and the temperature is 2SC. (The Kp for the reaction is 4.0 x1031 at this temperature.) ×10 atm

Explanation / Answer

ANSWER

N2 (g) + O2 (g)<--------> 2NO (g)

Kp = P2(NO) / P(O2) P(N2)

P(NO) = partial pressure of NO

P(O2) = partial pressure of O2

P(N2) = partial pressure of N2

4.5 X 10-11 = P2(NO) / 3.0 X 0.012

P2(NO) = 4.5 X 10-11 X 3.0 X 0.012 = 1.62 X 10-12

P(NO) = (1.62 X 10-12)1/2 = 1.27 X 10-6 atm

(b) Kp = P2(NO) / P(O2) P(N2)

4.0 X 10-31 = P2(NO) / 0.78 X 0.21

P2(NO) = 4.0 X 10-31 X 0.78 X 0.21 = 6.55 X 10-32

P(NO) = (6.55 X 10-32)1/2 = 2.56 X 10-16 atm

(c) The reaction is ENDOTHERMIC. Equilibrium constant increases with increase in temperature and decreases with decrease in temperature if the FORWARD REACTION IS ENDOTHERMIC.

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