At 430 degree C, the equilibrium constant(k_p) for the reaction 2NO(g)+O_2(g) 2N

Question

At 430 degree C, the equilibrium constant(k_p) for the reaction 2NO(g)+O_2(g) 2NO_2(g) is 1.5 times 10^5. In one experiment, the initial pressures of NO, O_2 and NO_2 are 8.0 times 10^-3 atm,9.9 times 10^-2 atm, and 0.21 atm, respectively. Calculate Q_p and predict the direction that the net reaction will shift to reach equilibrium. What is Q_p for the experiment? Times 10 In which direction will the system proceed to reach equilibrium? The reaction will proceed to the left. The reaction is at equilibrium. The reaction will proceed to the right.

Explanation / Answer

At 430 ^c The equilibrium consatnt K p for the reacion is 1.5*10^5

The relation between equilibrium consatnt K p and reaction quotient Q as follows:

Q = K

When Q=K, the system is at equilibrium and there is no shift to either the left or the right.

Q < K

When Q<K, there are more reactants than products. As a result, some of the reactants will become products, causing the reaction to shift to the right.

Q > K

When Q>K, there are more products than reactants. To decrease the amount of products, the reaction will shift to the left and produce more reactants. For Q>K:

First calculate the Q for this reaction as follows:

Qp = p(NO2^2) / (p(NO^2) x p(O2))

Qp = (0.21^2) / ((8.0 x 10^-3)^2)) x ( 9.9 x 10^-2)

Qp =0.0441 / 6.336*10^-6

Qp = 6960.23

Qp =6.9*10^3

Since Qp; 6.9*10^3 is < kp; 1.5*10^5

, the reaction will shift towards the right (to the products) to reach equilibrium.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.