This question has multiple parts. Work all the parts to get the most points. The

Question

This question has multiple parts. Work all the parts to get the most points. The reaction 2 NO(g) + 2 H_2(g) rightarrow N_2(g) + 2 H_2O(g) was studied at 904 degree C, and the data in the table were collected. Determine the order of the reaction for each reactant. Determine the order of the reaction for each reactant. Write the rate equation for the reaction. What is the rate constant for the reaction? Find the rate of appearance of N_2 at the instant when [NO] = 0.380 mol/L and [H_2] = 0.280 mol/L.

Explanation / Answer

Answer – Given the data of the concentrations of both reactant and rate of formation.

Order of reaction – assume rate law

Rate = k [NO]m [H2]n

The m and n order with respect to NO and H2, so rate law are as follow -

Rate1 = k [NO]1m [H2]1n

Rate2 = k [NO]2m [H2]2n

Rate 3 = k [NO]3m [H2]3n

Rate 4 = k [NO]4m [H2]4n

Now first we need to calculate order of NO

So,

Rate2/ Rate1 = k [NO]2m [H2]2n / k [NO]1m [H2]1n

0.0371 / 0.147 = (0.210)m /(0.420)m * (0.133)n / (0.133)n

0.25 = (0.5)m

So, m = 2

So order with respect to NO is 2 order

Now order of H2

Rate3/Rate2 = k [NO]3m [H2]3n / k [NO]2m [H2]2n

0.0741 / 0.0371 = (0.210)m /(0.210)m *(0.266)n /(0.133)n

2 = (2)n

So, n = 1

So order with respect H2 is 1

overall order =2 + 1 = 3

So rate law, Rate = k [NO]2 [H2]

So overall order is third order

Now we need to calculate the rate constant

We know rate law

Rate = k [NO]2 [H2]

0.148 M/s = k *(0.420 M)2(0.133M)

k = 0.148 M.s-1/ 0.0235 M3

= 6.31 M-2. s-1

d) [NO] = 0.380 M , [H2] = 0.280 M

rate = 6.31 M-2. s-1 * (0.380M)2 *(0.280 M)

       = 0.255 M/s

Rate of appearance = rate of disappearance

Rate = - ½ [NO]/ t = - ½ [H2]/ t = [N2]/ t

So, rate of appearance N2

[N2]/ t = ½ *0.255 M/s

                = 0.127 M/s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.