NETICS AND EQUIIBROUM = Using a rate law The rate of a certain reaction is given
ID: 1087225 • Letter: N
Question
NETICS AND EQUIIBROUM = Using a rate law The rate of a certain reaction is given by the following rate law rate=k[N2]2[H2]2 Use this information to answer the questions below what is the reaction order in N2? What is the reaction order in H2? what is overall reaction order? At a certain concentration of N2 and H2, the initial rate of reaction is 3.0 x 10 M/s. What would the initial rate of the reaction be if the concentration of N2 were halved, Be sure your answer has the correct number of significant digits The rate of the reaction is measured to be 77.0 M/ s when [N2] 1.0 M and [H2) 1.3 M. Calculate the value of the rate constant. Be sure your answer has the correct number of -3-1 | k-uM significant digits.Explanation / Answer
Answer:
Given rate=k[N2]^2[H2]^2
=> The order of the reaction with respect to N2 is 2.
=> The order of the reaction with respect to H2 is also 2.
=> The overall order=order in N2+order in H2=2+2=4.
=> Given intial rate=3×10^5 M/s=k[N2]^2[H2]^2
If [N2]=halved=[N2]/2
Then initial rate=k[N2/2]^2[H2]^2=(k[N2]^2[H2]^2)/4=(3×10^5 M/s)/4
Initial rate when [N2] is halved=7.5x10^4 M/s.
=> Given rate of reaction=77.0 M/s, [N2]=1.0 M anf [H2]=1.3 M.
Rate=k[N2]^2[H2]^2
77 M/s=k(1.0 M)^2 x (1.3 M)^2
k=(77 M/s)/(1.69M^4)
k=45.56 M^-3 s^-1.
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