The first step in industrial nitric acid production is the catalyzed oxidation o

Question

The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different reaction predominates: 4NH_3(g) + 3O_2(g) = 2N_2(g) + 6H_2O(g) When 0.0220 mol gaseous NH3 and 0.0140 mol gaseous O_2 are placed in a 1.00 L container at a certain temperature, the N_2 concentration at equilibrium is 1.30 times 10^3 M. Calculate K_eq for the reaction at this temperature. 92.4 Start by setting up a reaction table. You'll also want to write the expression for the equilibrium constant. Remember that we write equilibrium constants with no units. Tries 1/99 Previous Tries

Explanation / Answer

Since volume is 1 L, number of moles will be equal to concentration

4NH3 (g) + 3O2 (g) <----> 2N2 (g) + 6H2O (g)
0.0220 0.0140 0 0 (initial)
0.0220-4x 0.0140-3x 2x 6x (at equilibrium)

[N2] = 2x = 1.30*10^-3 M
x = 6.50*10^-4 M

Kc = [N2]^2 [H2O]^6 / {[NH3]^4 * [O2]^3}
= (2x)^2 * (6x)^6 / {(0.0220-4x)^4 * (0.0140-3x)^3}
= (2*6.50*10^-4)^2 * (6*6.50*10^-4)^6 / {(0.0220-4*6.50*10^-4)^4 * (0.0140-3*6.50*10^-4)^3}
= (1.69*10^-6) *(3.52*10^-15) / {1.42*10^-7 * 1.75*10^-6}
= 2.39*10^-8
Answer: 2.39*10^-8

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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