Part 1) 2NO(g)+2H2(g)N2(g)+2H2O(g) The rate law for this reaction is first order
ID: 888687 • Letter: P
Question
Part 1)
2NO(g)+2H2(g)N2(g)+2H2O(g)
The rate law for this reaction is first order in H2 and second order in NO. Write the rate law.
The rate law for this reaction is first order in and second order in . Write the rate law
Part 2)
If the rate constant for this reaction at 1000 K is 6.0×104M2s1, what is the reaction rate
when [NO]=3.60×102M and [H2]= 1.10×102M?
Express your answer using two significant figures.
What is the reaction rate at 1000 K when the concentration of NO is increased to 0.14 M, while the concentration of H2 is 1.10×102M?
Express your answer using two significant figures.
rate=k[H2]
rate=k[H2][NO] rate=k[H2][NO]2 rate=k[H2]2[NO]2Part 2)
If the rate constant for this reaction at 1000 K is 6.0×104M2s1, what is the reaction rate
when [NO]=3.60×102M and [H2]= 1.10×102M?
Express your answer using two significant figures.
What is the reaction rate at 1000 K when the concentration of NO is increased to 0.14 M, while the concentration of H2 is 1.10×102M?
Express your answer using two significant figures.
rate=k[H2]
Explanation / Answer
Solution :-
Part 1 ) The reaction is first order with respect to H2 second order with respect to NO therefore
Rate law for the reaction is
Rate = k [H2] [NO]^2
Part 2
If the rate constant for this reaction at 1000 K is 6.0×104M?2s?1, what is the reaction rate
when [NO]=3.60×10?2M and [H2]= 1.10×10?2M?
We know the rate law
Rate = k [H2][NO]^2
Lets put the values in the formula and calculate the rate
Rate= 6.0×10^4M?2s?1 [1.10×10?2] [3.60×10?2]^2
Rate = 8.6*10^-1 M-1 s-1
part 3 ) What is the reaction rate at 1000 K when the concentration of NO is increased to 0.14 M, while the concentration of H2 is 1.10×10?2M?
Solution :-
Rate law is
Rate = k[H2][NO]^2
Lets put the values in the formula
Rate = 6.0*10^4 M-2.s-1 [1.10*10^-2][0.14]^2
Rate = 12.9 M-1.s-1