I need help with Chemistry Question no.10. Tks. prob0001pdf Adobe Acrobat Reader

Question

I need help with Chemistry Question no.10. Tks.

prob0001pdf Adobe Acrobat Reader DC File Edit View Window Help Sign In Export PDF at 45 °C. a. How long does it take for the quantity of N2O5 to be reduced to 2.5 g? b. How many liters of O2, measured at 745 mm Hg and 45 °C are produced up to this point? A certain reaction has the following general format At a certain temperature and [A] 0.0200 MI, experimental data was collected for this reaction A plot of InA] versus time resulted in a straight line with a slope value of -2.97 x 102 min1. a. Determine the rate law, the integrated rate law, and the value of k for this reaction. b. Calculate the half life for this reaction c. How much time is required for the concentration of A to decrease to 2.50 x 10 M? Create PDF v Edit PDF Combine Files ^ Adobe Acrobat Pro DC Combine two or more files into a single POF Learn more organize Pages 10.Acertain reaction has the following general fornm aA bB Protect At a certain temperature and [A]°-2.80 x 10M, experimental data was collected for the reaction. A plot of 1/[A] versus time resulted in a straight line with a slope value of 3.60 x 10 L/(mol s). a. Determine the rate law, the integrated rate law, and the value of k for this reaction b. Calculate the first half life for this reaction c. How much time is required for the concentration of A to decrease to 7.00 x 10 M? Optimize PDF Fill& Sign Send for Signature Send & 1 rack Store and share files in the Document Cloud Learn More 1054 PM Type here to search e OChegg Study | Guid. aprob0001.pdf . Ado ^ 4x ENG 2018-01-12

Explanation / Answer

Answer – We are given a chemical reaction and we need to determine first whether it is a zero, first, or second order reaction.

We are given that graph 1/[A] versus time gives straight line with positive slope and from the we can get whether it is a second order reaction, since for second order of reaction 1/[A] versus time is a straight line with k = slope of the line. There is also given slope which equal to rate constant value with unit L/ (mol.s) and from that also there is confirm it is second order of reaction.

Given values, Slope = 3.60 x 10-2 L/(mol.s) , [A0] = 2.80 x 10-3 M

Reaction – a A ----> b B

a)As per the above explanation we know there is second order of reaction, so

Rate law = k [A]2

Integrated rate law for second order of reaction –

1/ [A] = kt + 1/ [A0]

We know for second order of reaction slope which equal to rate constant value, so

Rate constant , k = 3.60 x 10-2 L/(mol.s)

b) We know half life formula for the second order of reaction –

t ½ = 1 / k x [A0]

t ½ = 1 / 3.60 x 10-2 L.mol-1.s-1 * 2.80 x 10-3 M

       = 9.920 x 103 s

Th first half life for this reaction is 9.920 x 103 seconds.

c) We are given , [A] = 7.00 x 10-4 M and we need to calculate the time.

WE know the initial concentration, [A0] = 2.80 x 10-3 M and it is second order of reaction, so using the integrated rate law for second order of reaction-

1/ [A] = kt + 1/ [A0]

1 / 7.00 x 10-4 M = 3.60 x 10-2 L.mol-1.s-1 *t + 1/ 2.80 x 10-3 M

1428.5 = 3.60 x 10-2 L.mol-1.s-1 *t + 357.14

3.60 x 10-2 L.mol-1.s-1 *t = 1428.5 – 357.14

                                      = 1071.4

So, t = 1071.4 / 3.60 x 10-2

       = 2.98 x 104 seconds

The time required for the concentration of A to decrease 7.00 x 10-4 M is 2.98 x 104 seconds

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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