For the reaction 6CH_2O (aq) + 4NH_3 (aq) rightarrow (CH_2)_6N_4 (aq) + 6H_2O(I)

Question

For the reaction 6CH_2O (aq) + 4NH_3 (aq) rightarrow (CH_2)_6N_4 (aq) + 6H_2O(I) the rate of the reaction may be expressed as -1/6 times delta [CH_2O]/delta t. What is an equivalent expression for the rate of the reaction? A) 1/2 times delta [(CH_2)_6N_4/delta t B) 6 times delta [CH_2O]/delta t C) -6 times delta [CH_2O]/delta t D) 1/6 times delta [H_2O]/delta t E) -1/6 times delta [H_2O]/delta t For the first-order reaction 1/2 N_2O_4 (g) rightarrow NO_2 (g): delta H = 28.6 kJ the rate constant is k = 2.95 times 10^5 s^-1 at -8 degree C, and the activation energy is 53.7 kJ/mol. What is the rate constant at 6 degree C? A) 5.55 times 10^5 s^-1 B) 1.00 times 10^6 s^-1 C) 2.95 times 10^5 s^-1 D) 3.11 times 10^5 s^-1 E) 4.49 times 10^4 s^-1 Exactly 1.0 mol N_2O_4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N_2O_4 (g) doubleheadarrow 2NO_2 (g). If at equilibrium the N_2O_4 is 37.0% dissociated, what is the value of the equilibrium constant, K_c, for the reaction under these conditions? A) 1.17 B) 0.87 C) 1.2 D) 0.55 E) 0.22

Explanation / Answer

7. Given the reaction, the rate can be represented by,

D)

8. Using,

ln(k2/k1) = Ea/R[1/T1 - 1/T2]

ln(k2/2.95 x 10^5) = 53700/8.314 [1/265 - 1/279]

rate constant at 6 oC would be k2 = 1.00 x 10^6 s-1

Answer, B. 1.00 x 10^6 s-1

9. With 37% dissociated,

N2O4 remaining = 1 - 0.37 = 0.63 mol/L

NO2 formed = 2 x 0.37 = 0.74 mol/L

Equilibrium constant Kc,

Kc = (0.74)^2/0.63 = 0.87

Answer : B) 0.87

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

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