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The reaction: 2NO (g) + 2H2 (g)  N2 (g) + 2H2O (g) a. Determine the rate...

The reaction: 2NO (g) + 2H2 (g)  N2 (g) + 2H2O (g)

a. Determine the rate order for the reaction.

b. Write a rate law, including the value for k.

[NO] [H2] Rate M/s0.

420 0.122 0.136

0.210 0.122 0.0339

0.210 0.244 0.0678

0.105 0.488 0.0339

Solutions

Expert Solution

Let the rate law be,

Rate law = k[NO]^x[H2]^y

From the table,

0.136 = k[0.420]^x[0.122]^y .....(1)

0.0339 = k[0.210]^x[0.122]^y.....(2)

0.0678 = k[0.210]^x[0.244]^y ......(3)

Dividing eq1 by eq2 we get,

0.136/0.0339 = [0.420/0.210]^x

4.0 = [2]^x

[2]²=[2]^x

x = 2

Now dividing eq3 by eq2, we get

0.0678/0.0339 = [0.244/0.122]^y

2 = [2]^y

[2]^{1 =} [2]^y

y = 1

So, order of the reaction is x+y = 2+1 = 3

Rate law = k[NO]²[H2]¹

Lets find out the value of k,

From eq1,

k = 0.136/[0.420]²[0.122]¹

= 0.136/0.0215

= 6.30

From eq2,

k = 0.0339/[0.210]²[0.122]¹

= 0.0339/0.00538

= 6.30

From eq 3,

k = 0.0678/[0.210]²[0.244]¹

= 0.0678/0.01076

= 6.30

Thus, k = 6.30

Hence rate law,

Rate = 6.30[NO]²[H2]¹

queen_honey_blossom answered 2 years ago

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