Answer:
Order respect to NO a=2
Order respect to H_{2} b=1
Rate constant k=250.1/M^{2}s
Explanation:
The law of velocity can be expressed by the equation
[tex]-r_{A} =k.[NO]^{a} .[H_{2}]^{b}[/tex]
For each experiment, we can write
[tex](1)1,3.10^{-5} =k.0,0050^{a} .0,0020^{b}[/tex]
[tex](2)5,0.10^{-5} =k.0,0100^{a} .0,0020^{b} [/tex]
[tex](3)10,0.10^{-5} =k.0,0100^{a} .0,0040^{b} [/tex]
making the quotient between (1) and (2) we obtain
[tex](1,3.10^{-5}/5,0.10^{-5} )=(0,0050/0,0100)^{a} [/tex]
We get the value of the order of reaction respect to NO a≈2 so the rate of the reaction gets quadruplicated when [NO] duplicates
Doing the same between 2 and 3
[tex](5,0.10^{-5}/10,0.10^{-5})=(0,0020/0,0040)^{b} [/tex]
We got the value of b=1 so the rate gets duplicated when [H_{2}] duplicates
Then, for any experiment we can calculate the rate constant k by the first equation . e.g. for the second experiment
[tex]5,0.10^{-5} =k.0,0100^{2} .0,0020^{1} [/tex]
Then
[tex]250\frac{1}{M^{2}s}[/tex]
Answer:
Rate = 1.23*[NO]^2*[H2]
Explanation:
Data from question is unintelligible, so I'm going to use the table attached (the procedure doesn't change).
From experiments 1 and 2, when [H2] doubles, rate doubles; then, the order of reaction for [H2] is 1.
From experiments 2 and 3, when [NO] doubles, rate quadruples; then, the order of reaction for [NO] is 2.
The formula for the rate law is:
Rate = k*[NO]^2*[H2]
solving for the rate constant and replacing with the initial rate we get:
k = initial rate/([NO]^2*[H2])
k = 1.23*10^-3/(0.1^2*0.1) = 2.46*10^-3/(0.1^2*0.2) = 4.92*10^-3/(0.2^2*0.1) = 1.23
