Question

The reaction of nitric oxide with hydrogen at 1280 C is 2NO(g) +2H2(g)N2(g)+2H2O(g) From the following data collected at this temperature, determine the rate law and calculate the rate constant. Initial rate (M/s) 1.3 x 10 5.0 x 10 Experiment 1 2 3 NO 0050 0100 0100 .0020 .0020 0040 10.0 x 105

Answer 1

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

Answer 2

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

$-r_{A} =k.[NO]^{a} .[H_{2}]^{b}$

For each experiment, we can write  

$(1)1,3.10^{-5} =k.0,0050^{a} .0,0020^{b}$

$(2)5,0.10^{-5} =k.0,0100^{a} .0,0020^{b}$

$(3)10,0.10^{-5} =k.0,0100^{a} .0,0040^{b}$

making the quotient between (1) and (2) we obtain

$(1,3.10^{-5}/5,0.10^{-5} )=(0,0050/0,0100)^{a}$

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  

$(5,0.10^{-5}/10,0.10^{-5})=(0,0020/0,0040)^{b}$

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

$5,0.10^{-5} =k.0,0100^{2} .0,0020^{1}$

Then

$250\frac{1}{M^{2}s}$