CHM 111 Exam 3 | Module 3 Fall 2017 Nitrogen(II) oxide reacts with chlorine acco

Question

CHM 111 Exam 3 | Module 3 Fall 2017 Nitrogen(II) oxide reacts with chlorine according to the equation: 2NO(g) + Cl2(g)- 2NOCI(g) The following initial rates of reaction have been observed for certain reactant concentrations: [Clyl (mol/L) Rate (mol/L/h) 0.50 0.50 1.00 INO] (mol/L 0.50 1.00 1.00 1.14 4.56 9.12 10. What is the rate equation that describes the rate's dependence on the concentrations of NO and Clz? 11. 12. What is the rate constant? What are the orders with respect to each reactant? Butane exists as two isomers, n-butane and isobutane. CH2 CH n-butane Kr= 2.5 at 25 °C isomers at equilibrium with a total pressure of 1.22 atm? Given the following information: 13. What is the pressure of isobutane in a container of the two [NH3] = 0.20 M, [N2] = 1.00 M, [H2] = 1.00 M 14. 15. Calculate the reaction quotient Determine the direction in which each system will proceed to reach equilibrium.

Explanation / Answer

In order to calculate the rate law expression for a A+B reaction, we need to apply Initial Rates Method.

Note that the generic formula goes as follows:

r = k [A]^a [B]^b

Note that if we got at least 3 sets of point, in which we have A and B constant, then we could use:

r1 / r2 = (k1 [A]1^a [B]1^b) / (k2 [A]2^a [B]2^b)

If we assume K1 and K2 are constant, then K1= K2 cancel each other

r1 / r2 = ([A]1^a [B]1^b) / ( [A]2^a [B]2^b)

Then, order according to [A] and [B]

r1 / r2 = ([A]1/[A2])^a * ([B]1/[B]2)^b

If we get two points in which A1 = A2, then we could get B, and vise versa for A...

From the data shown in YOUR table

choose point 1 and 2 so Cl2 cancels

(1.14/4.56) = 0.5/1^a

a = ln((1.14/4.56)) /ln(0.5) = 2

2nd order with respec tto NO

then,

choose point 2 and 3

4.56/9.12 = (.5/1)^b

b = ln(0.5) /( ln(0.5) = 1

1st order with repect to Cl2

Rate = k*[NO]^2[Cl2]

choose any point for k

k = Rate /([NO]^2[Cl2])

k = (1.14)/((0.5^2)(0.5)) =

K = 9.12

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