The rate constant for a second-order reaction is 0.13 M^-1 s^-1. If the initial
ID: 1067932 • Letter: T
Question
The rate constant for a second-order reaction is 0.13 M^-1 s^-1. If the initial concentration of reactant is 0.26 mol/L, it takes s for the concentration to decrease to 0.11 mol/L. 0.017 0.68 40 9.1 5.2 For the elementary reaction NO_3 + CO rightarrow NO_2 + CO_2 the molecularity of the reaction is, and the rate law is rate = 4, k[NO_2][CO_2]/[NO_3][CO] 4, k[NO_3][CO][NO_2][CO_2] 2, k[NO_2][CO_2] 2, k[NO_3][CO]/[NO_2][CO_2] 2, k[NO_3][CO] Which of the following expressions is the correct equilibrium-constant expression for the reaction below? CO_2(s) + H_2O (I) rightarrow H^+ (aq) + HCO_3^-(aq) [H^+][HCO_3^-] [H^+][HCO_3^-]/[CO_2][H_2O] [CO_2][H_2O]/[H^+][HCO_3^-] [H^+][HCO_3^-]/[CO_2] How is the reaction quotient used to determine whether a system is at equilibrium? The reaction quotient must be satisfied for equilibrium to be achieved. At equilibrium, the reaction quotient is undefined. The reaction is at equilibrium when Q = K_eq. The reaction is at equilibrium when Q > K_eq. The reaction is at equilibrium when QExplanation / Answer
1.
The rate constant, K= 0.13/m.sec
For a second order reaction, 1/CA= 1/CAO+ Kt
CA= concentration at any time, t = 0.11 mol/L
CAO= concentration at any time, t=0, is 0.26 mol/L
1/0.11= 1/0.26+0.13*t
Hence time taken, t= 40.34 seconds. ( C is the correct answer)
2.
Molecularity refer to number of molecules participating in the reaction. So there are two molecules (NO3, CO). Their coefficients are 1, 1. So total moleculartiy is 2.
For elementary reaction, there will be correspondence between stoichiometry and rate expression. Accordingly, Rate= K[NO3][CO]. K is rate constant. Hence Correct answer is E.
3. For the reaction, CO2 is solid and H2O(l) is a liquid and their activity become 1.
Hence Equilibrium constant = K[ H+] [HCO3-]/ [CO2] [H2O] = K[ H+][HCO3-]
4. When the reaction becomes equal to K ( equilibrium constant), the system is said to be in Equilibrium.
for Q>K, the system shifts towards reactants side and for Q<K, the reaction shifts to the product side.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.