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baherus [9]
3 years ago
14

What would happen to the rate of a reaction with rate law rate = k [NO]2[Hz] if

Chemistry
1 answer:
leonid [27]3 years ago
8 0
The rate of a reaction with this rate law would increase by a factor of 4 if NO concentration were doubled.
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Switch the coordinates and change the sign of the second one by multiplying it by negative 1.

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Here are some examples and a more general way to understand the problem.

Consider the point (1,1), a 90 degree rotation clockwise about the origin would move it into the 4th quadrant.

The new point is (1,-1) , similarly (-4,2)-> (2,4), (-4,3)-> (3,4)

We take a point p= (x,y) the the result of rotation p 90 clockwise about the orgin is a new point p'=(x',y')= (-y, x). .

In the case of p=(1,0) the new point is p'= (0, -1)

One can use a matrix where the first row is cos(a), sin(a) and the second row is

-sin(a) cos(a) for any clockwise rotation of a degrees about the origin.

If we let a=90 degrees we have

[0 1] as the first row and [-1 0] as the second row. So the matrix is:

|0 1|

|-1 0|

Call that matrix M

So a point p= (x,y) can be multiplied by M as follows Mp=p' where p' is the rotated point.

If p=(-4,2) then Mp

is M(-4,2) which after matrix multiplication means x'=0*-4+1*2=2 and y'=-1*-4+0*2=4

So p'=(2,4)

Try it with (1,0)

x'=1*0+0*1=0

y'=-1*1+0*1=-1

so p'=(0,-1) and (1,0)->(0,-1)

How about the point on the y axis (0,1), it should go to the point (1,0)

0*1+1*1=1 and -1*0+0*1 gives you the pont (1,0) ( we don't see the negative sign because -0 is just 0)

7 0
3 years ago
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