Question

Consider the following reaction:
2 NO(g) + 2 H2(g)-->N2(g) + 2 H2O(g)
(a) The rate law for this reaction is first order in H2 and
second order in NO. Write the rate law. (b) If the rate
constant for this reaction at 1000 K is 6.0 * 104 M-2 s-1,
what is the reaction rate when 3NO4 = 0.035 M and
3H24 = 0.015 M? (c) What is the reaction rate at 1000 K
when the concentration of NO is increased to 0.10 M,
while the concentration of H2 is 0.010 M? (d) What is the
reaction rate at 1000 K if [NO] is decreased to 0.010 M and
3H24 is increased to 0.030 M?

Answer 1

a) If the reaction is first order with H2 and second order with NO, then this means that rate law will be:

r = k [H₂]¹ [NO]²

where r = rate

k = rate constant

The concentrations will be raised to the order given for each reactant.

b) rate constant = k = 6.0*10⁴ M^-2 s^-1

[NO] = 0.035 M

[H2] = 0.015 M

So, according to rate law given above, rate will be:

r = k [H₂]¹ [NO]²

= 6.0*10⁴ [0.015]¹ [0.035]²

= 1.10 M s^-1

c) rate constant = k = 6.0*10⁴ M^-2 s^-1

[NO] = 0.10 M

[H2] = 0.010 M

So, according to rate law given above, rate will be:

r = k [H₂]¹ [NO]²

= 6.0*10⁴ [0.010]¹ [0.10]²

= 6.0 M s^-1

d) rate constant = k = 6.0*10⁴ M^-2 s^-1

[NO] = 0.010 M

[H2] = 0.030 M

So, according to rate law given above, rate will be:

r = k [H₂]¹ [NO]²

= 6.0*10⁴ [0.030]¹ [0.010]²

= 0.18 M s^-1