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solong [7]
3 years ago
12

1. A scientist studies the reaction 2NO2(g) 2NO(g) + O2(g). She performs three experiments using different concentrations of NO2

and measures the initial reaction rate.
Experiment : [NO2] (mol/L) : Initial Rate ((mol/L)/s)
1 : 0.1 : 0.006
2 : 0.3 : 0.054
3 : 0.5 : 0.150
A. What is the ratio of the concentrations between Trials 1 and 2? (2 points)

B. What is the ratio of the initial reaction rates between Trials 1 and 2? (2 points)

C. What is the exponent for [NO2] in the rate law? (2 points)

D. Write the rate law. (2 points)

E. Solve for the value of k. (2 points)

F. What is the overall reaction order? (2 points)
Chemistry
1 answer:
DIA [1.3K]3 years ago
6 0
A. The concentration ratio between trials 2 and 1 is: 0.3 / 0.1 = 3.
B. The ratio between reaction rates is: 0.054 / 0.006 = 9.
C. We solve the equation: (concentration ratio)^x = (reaction rate ratio): 3^x = 9, which gives x = 2, and the exponent in the rate law is 2.
D. The rate law will have the equation: r = k[NO2]^2
E. To solve for k, we can substitute the values for [NO2] = 0.1, rate = 0.006. This gives: 0.006 = k(0.1)^2, which yields k = 0.6 L/mol-s.
F. Since the exponent is 2, this is a second-order reaction.
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3 years ago
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What is the concentration of a solution containing 1.11 g sugar (sucrose, C12H22O11, MW = 342.3 g/mol, d = 1.587 g/cm3) in 432 m
djyliett [7]

Answer:

0.0075 M

0.0060 m

Explanation:

Our strategy here is to use the definition of molarity and molality to solve this question.

The molarity is the number of moles of solute, sucrose in this case, per liter of solution.

The molality is the number of moles of solute per kilogram of solvent.

So the molarity of the  solution is

M = moles of solute/ V solution

As we see we need the volume of solution since we are only given the volume of solvent, but this will be easy to compute since we have the density of  sucrose.

So determine the moles of sucrose , and the volume of solution:

Moles sucrose = 1.11 g/342.3 g/mol = 3.24 x 10⁻³ M

Volume of solution = Vol Sucrose + Vol glycerine

d = m/V ⇒ Vsucrose = m / d = 1.11 g/ 1.587 g/cm³ = 0.70 cm³

Vol solution = 432 mL + 0.70 mL = 432.7 mL  (1cm³  = 1 mL)

Vol solution = 432.7 mL x 1 L / 1000 mL = 0.4327 L

⇒ M = 3.2 x 10⁻³  mol / 0.4327 L = 0.0075  M

For the molarity what we need is to first calculate the kilograms of glycerine from the given density:

d = m/v ⇒ m = d x v = 1.261 g/cm³ x  432 cm³ = 544.75 g

Converting to Kg:

544.75 g x 1 Kg/ 1000 g = 0.544 kg

Now the molality is

m = mol sucrose/ kg solvent = 3.24 x 10⁻³ mol / 0.544 Kg = 0.0060 m

Note: In the calculation for  volume of solution we could have approximated it to that of just glycerine, but since the density of sucrose was given we calculated the total volume of solution to be more rigorous.

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