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Please help I know its a lot. Will rate!! 50. Questions 50-54 refer to the follo

ID: 571919 • Letter: P

Question

Please help I know its a lot. Will rate!!

50. Questions 50-54 refer to the following information. CHy H,C CHj cis-2-butene trans-2-butene The half-life (bs) of the catalyzed isomerism of cis-2-butene gas to produce trans-2-butene gas represented above, was measured under various conditions, as shown in the table below. Trial Number Initial Pais2-utee(torr) LT(K) (s) 300. 600. 300. 150. 3.00 6.00 9.00 10.0 350. 100 350. 100. 50. 50. 365 365 Which trial had the highest initial mol amount of cis-2-butene? (A) Trial 1 (B) Trial 2 (C) Trial 3 (D) Trial 4 AP Chem Exam 2014 FROR 51. The reaction is considered to be first order. Which of the following statements correctly interprets the data to justify this claim? (A) As the temperature increases, the half-life decreases because the rate constant, k (B) Comparing trials 1 and 2, when the initial pressure is changed from 300. torr to 600. torr, the (C) Comparing trials 1 and 2, when the volume increases from 3.00 L to 6.00 L, the half-life (D) Comparing trials 1 and 4, when the initial pressure is halved from 300. torr to 150. torr, the proportionally increases. half-life remains constant at 100. s. remains constant. half-life is also halved from 100. s to 50. s AP Chea Exom 2014 FRO#7 Which of the following correctly explains the reduction in half-life between trial 2 and trial 3? (A) There is a larger percentage of cis-2-butene molecules with sufficient kinetic energy to 52. (B) The activation energy, Ea, is lower at 365 K than at 350. K; therefore a greater percentage of (C) At higher temperatures the sp2 hybridization of cis-2-butene molecules is more easily (D) The cis-2-butene has greater repulsion between the methyl groups at higher temperatures; overcome the activation energy for the reaction, thus the rate increases. the cis-2-butene molecules can be converted to trans-2-butene molecules. changed to the sp hybridization of trans-2-butene. therefore the forward reaction rate increases with temperature.

Explanation / Answer

Ans 50

From the ideal gas equation

Moles n = PV /RT

For first trial

Pressure P = 300 torr x 1 atm/760torr = 0.3947 atm

Volume V = 3 L

Temperature T = 350 K

Gas constant R = 0.08206 L-atm/mol-K

Moles n = (0.3947 x 3)/(0.08206 x 350)

n = 0.04122 mol

Similarly for trial 2

n = 0.1649 mol

Trial 3

n = 0.1186 mol

Trial 4

n = 0.06588 mol

For trial 2, moles of cis 2 butene are highest

Option B is the correct answer

Ans 51

For first order reaction

Half life t1/2 = 0.6932/k

Rate constant k depends on the temperature, as temperature increases, rate constant also increases and increasing the rate constant, the half-life will decrease because half life is inversely proportional to the rate constant

Option A is the correct answer

Ans 52

Half life reduction means increase in the rate constant and increase in the reaction rate.

Higher the kinetic energy, higher the rate constant, higher the reaction rate

Option A is the correct answer

Ans 53

For trial 1

Half life

t1/2 = 0.6932/k

k1 = 0.6932/100 = 0.006932 s-1

Trial 2

k1 = k2 = 0.006932 s-1

For trial 3

k3 = 0.6932/50 = 0.13864 s-1

Rate law for first order reaction

r = k [A]

For constant concentration let's say for unit concentration

r = k

k1 = k2 < k3

r1 = r2 < r3

Option B is the correct answer

Ans 54

There is a mistake in this question. It would be 350 K not 360.

At 350 K

Half life

t1/2 = 0.6932/k

k1 = 0.6932/100 = 0.006932 s-1

Option A is the correct answer

However I will also calculate the rate constant at 360 K

From Arrhenius equation

ln (k2/k3) = (E/R) (1/T3 - 1/T2)

ln (0.006932/0.13864) = (E/8.314)* (1/365 - 1/350)

E = 212162 J/mol

Again From Arrhenius equation

ln (k/k3) = (E/R) (1/T3 - 1/T)

ln (k/0.13864) = (212162/8.314)* (1/365 - 1/360)

ln (k/0.13864) = - 0.9710

k = 0.0525 s-1 ( at 360 K)

Neither option is matched.