Use the following data to determine the rate law for the reaction shown below. 2

Question

Use the following data to determine the rate law for the reaction shown below. 2NO + H_2 rightarrow N_2O + H_2O The half-life for a first order reaction is 45 min. What is the rate constant in units of s^-1? The following initial rate data apply to the reaction below. Which of the following is the rate law (rate equation) for this reaction? Which is the correct equilibrium constant expression for the following reaction? Calculate K_p for the reaction 2NOCl(g) 2NO(g) + Cl_2(g) at 400degree C if K_c at 400 degree C fot this reaction is 2.1 times 10^-2.

Explanation / Answer

1)Given chemical transformation:

2NO +H2 ---------> N2O + H2O

Experimental data:

Exp-#

[NO]

[H2]

Initial rate

1

0.021

0.065

1.46 M/min

2

0.021

0260 (i.e.4x0.065)

1.46 M/min

3

0.042(i.e. 2x0.021)

0.065

5.84 M/min

(i.e. 4x1.46)

A)From Exp-#-1 and 2, where [NO] is kept constant and [H2] is made 4 times but still initial rate remains constant hence its clear that initial rate is independent of [H2].

B)From Exp-#-1 and 3, where [NO] is doubled and [H2] kept constant and we can see that the initial rate is quadrupled (4 times) this mean initial rate is

Hence we can say that the rate law for this reaction is,

Rate [NO]2

Rate = K[NO]2.

=================================================================

2)For a First order reaction given that,

t½ = 45 min = 45 x 60 s = 270 s.

For first order reaction, half-life(t½) and rate constant(k) are related as,

k = ln(2) / t½

k =0.693 / t½………(ln(2) = 0.693)

k = 0.693/270

k = 2.6 x 10-3 s-1

Hence rate constant is 2.6 x 10-3 s-1

=======================================

3)Given transformation:

F2(g) + 2Cl2O(g) --------> 2FClO2(g) + Cl2(g)

Exp-#

[F2]

[Cl2O]

Initial rate(M/s)

1

0.05

0.010

5.0 x 10-4

2

0.05

0.040 (i.e. 4 x 0.010)

2.0 x 10-3

(i.e. 4 x 5.0 x 10-4)

3

0.10 (i.e. 2 x 0.05)

0.010

1.0x 10-3

(i.e. 2 x 10-4)

A) From exp-#-1 and 2, where [F2] kept constant and [Cl2O] is quadrupled by which initial rete is also quadrupled and hence we can say that, initial rate [Cl2O].

B)From exp-#-1 and 3, where [F2] is doubled and [Cl2O] kept constant and initial rate also observed to be doubled hence we can say that, initial rate [F2].

Hence combining these two we can write,

Rate [Cl2O][F2]

Rate = K[Cl2O][F2].

============================================================

4)Given transformation,

Fe2O3(s) +3H2(g) <---------> 2Fe (s) + 3H2O(g)

In equilibrium constant expression solid and liquid species do not appear and only gaseous species appears. And hence equilibrium constant expression is given as,

Kc = [H2O]3 / [H2]3

====================

5) Given transformation and data

2NOCl(g) < ------- > 2NO(g) + Cl2(g)

i.e. No. or moles of gaseous reactant = 2

No. of moles of gaseous product = 2 + 1 = 3

Hence, n = No. of moles of gaseous product – No. of moles of gaseous reactant

n = 3 – 2

n = 1.

Relation between Kp and Kc is

Kp = Kc x (RT)n.

With n = 1

Hence above equation takes form,

Kp = Kc x R x T……..(1)

We are given with, Kc = 2.1 x 10-2 at T = 400 0C = 400 + 273.15 = 673.15 K, R = 0.082 L-atm.K-1.mol-1.put all data in eq. 1

Kp =2.1 x 10-2 x 675.15 x 0.082

Kp =1.159

Kp = 1.2 (apprx.)

Hence Kp = 1.2.

=======================

Exp-#

[NO]

[H2]

Initial rate

1

0.021

0.065

1.46 M/min

2

0.021

0260 (i.e.4x0.065)

1.46 M/min

3

0.042(i.e. 2x0.021)

0.065

5.84 M/min

(i.e. 4x1.46)

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