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# Kinetics-rate chemistry help?

I am at a loss on these two questions and will gladly take only how to do them instead of just the answer.

1) The rate of the reaction

2A(g) + 3B(g) + 5C(g) =

is found to be independent of the concentration of A, first order in the concentration of B, and second order in the concentration of C. When 1.54 moles of A, 4.84 moles of B, and 3.05 moles of C are placed in a 2 liter container, the initial rate of the reaction is found to be 0.038 mol/liter· sec.

What is the numerical value of the reaction rate constant in units of moles, liters, and seconds?

Answer in units of M−2 · s−

2) The reaction

NO2(g) + CO(g) = NO(g) + CO2(g)

has been found to be second order with respect to NO2 and zero order with respect to CO. At a certain temperature, the rate constant is found experimentally to be 3.0 × 10−5 L mol · s

What is the rate of formation of CO2 at this temperature when the concentration of NO2 is 6.9 mol/L, that of CO is 6.2 mol/L, that of NO is 2.4 mol/L, and that of CO2 is 3.9 mol/L?

Answer in units of M· s−1

### 2 Answers

- GlenguinLv 78 years agoFavorite Answer
1) Based on what we are told, the rate law is:

Rate = k[B][C]².

Just calculate [B] and [C], substitute along with the rate, and solve for k.

2) Since this is a kinetics problem and not an equilibrium problem, all we care about is the concentration of the reactants. We are told that the reaction is zero order with respect to CO, so the rate law is simply:

Rate = k[NO2]².

Just substitute and calculate. This will tell us the rate of consumption of NO2. By the balanced equation, the rate of formation of CO2 will equal the rate of consumption of NO2.

- Anonymous4 years ago
For a known order technique the you need to use the equation: [At]=[Ai]e^-kt, log[At/Ai]=-kt/2.303 or ln[At/Ai]=-kt [At]=concentration at time t. [Ai]=preliminary concentration ok=value secure t=time Use whichever equation you have confidence maximum tender utilising. i will use [At]=[Ai]e^-kt because of the fact i think this is the least perplexing to utterly draw close. [At]=[Ai]e^-kt So shall we desirable down what all of us understand what we haven't have been given any theory: [At]=? [Ai]=0.800 M ok=2.48x10^-4/s t=10 min=600 s So we are fixing for [At]: [At]=0.800 M*e^(-2.48x10^-4/s*600 s)=0.6893 M it fairly is it!