Question

Use the data to determine rate law for reaction: 2NO+H2→N2O+H2O

Answer 1

Raw law Rate=k$[NO]^{x}$$[H_{2}] ^{y}$

In the above equation x is the order of the reaction with respect to NO

y is the order of the reaction with respect to $H_{2}$

we consider:

1)1.46M/min=k$(0.021)^{x}$$(0.065)^{y}$-------(1)

2)1.46M/min=k$(0.021)^{x}$$(0.260)^{y}$------(2)

3)5.84M/min=k$(0.042)^{x}$$(0.065)^{y}$------(3)

So Divide the 3rd reaction by the 1st we get the value of x

$\frac{5.84M/min}{1.46M/min}$=$\frac{k(0.042)^{x}(0.065)^{y}  }{k(0.021^{x}(0.065^{y}  }$

we get

4=$(2)^{x}$

$2^{2}$=$(2)^{x}$

x=2

so the order of the reaction with respect to NO is 2

similarly divide equation 2 with 1 we get

y=0

so the rate of law is=k$[NO]^{2}$$[H_{2}] ^{0}$=k$[NO]^{2}$