Nitrogen dioxide undergoes thermal decomposition according to the second-order reaction 2 NO₂ (g) $\rightarrow$2 NO(g) + O₂ .When 0.5000 M NO₂ is allowed to react for 9.00 minutes, its concentration falls to 0.0250 M.How long did it take for the concentration of NO₂ to fall to 0.2500 M?

Question

Quizplus · Accepted Answer

The rate law for the reaction is:Rate = k[NO2]^2The integrated rate law is:1/[NO2] - 1/[NO2]0 = ktwhere [NO2]0 is the initial concentration of NO2 and t is the time.We can use this equation to calculate the time it takes for the concentration of NO2 to fall to 0.2500 M:1/[0.2500 M] - 1/[0.5000 M] = k(t)k = (1/[0.2500 M] - 1/[0.5000 M]) / t = 0.00667 M^-1 min^-1Now we can use the integrated rate law to calculate the time it takes for the concentration of NO2 to fall to 0.2500 M:1/[0.2500 M] - 1/[0.5000 M] = k(t)t = (1/[0.2500 M] - 1/[0.5000 M]) / k = 0.474 minTherefore, the correct answer is A.

Nitrogen Dioxide Undergoes Thermal Decomposition According to the Second-Order Reaction →\rightarrow→

Nitrogen Dioxide Undergoes Thermal Decomposition According to the Second-Order Reaction $\rightarrow$