+ E 11. (14 pts) Given the following reaction and rate data CM 2A + B t c : Rate
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+ E 11. (14 pts) Given the following reaction and rate data CM 2A + B t c : Rate of Reaction (Ms) esperiment Al 0.080 0.16 0.130 0.380 | 0.300 0.380. -T0010 3 0,150 0.095 0.010 . 0.0050 4 0150 0.3800.070 0.080 (a) Determine the (differential) rate law. Show your work. (b) Determine the rate constant. (c) For Experiment 3, what is the rate of appearance of D? ACS-1. (1 pt) The solubility of a substance is 60 g per 100 mL of water at 15°C. A solution of this substance is prepared by dissolving 75 g in 100 mL. of water at 75°C. The solution is then cooled slowly to 15 C without any solid separating. The solution is (A) supersaturated at 75°C (C) unsaturated at 15°C (B) supersaturated at 15°C (D) saturated at 15°C ACS-2. (1 pt) A student wants to prepare 250. ml. of 0.10 M NaCl solution. Which procedure is most appropriate? (A) Add 5.84 g of NaCI to 250. mL. of H:0. (B) Add 1.46 g of NaCI to 250. ml. of H:O. (C) Dissolve 5.84 g of NaCI in 50 mL. of H:O and dilute to 250. mL (D) Dissolve 1.46 g of NaCl in 50 mL. of H:0 and dilute to 250. mL ACS-3. (1 pt) what is the initial rate of the reaction (in mo!1.sec-l) depicted by this graph? 0.3 0.2 0.I (A) 0.02 (B) 0.01 (C) 0.08 (D) 0.005 0 10 20 30 40 30 Time, s ACS-4, (1 pt) The rate law for the reaction A + B C + D is first order in [A] and second order in [B]. If [A] is halved and [B] is doubled, the rate of the reaction will (choose one answer): (A)remain the same (C) be increased by a factor of 4 (B) be increased by a factor of 2 (D) be increased by a factor of 8Explanation / Answer
11. (a) suppose the rate law is R= k [A]x [B]y [C]z
from set 1 and 2 we get R1/R2 = (0.15/0.30)x = 0.080/0.16
or, x= 1
from set 2 and 3 we will get, R2/R3 = (0.30/0.15)x . (0.38/0.095)y= 0.16/0.005
putting the value of x=1 we can get, (0.38/0.095)y= (0.16/0.005). 1/2 = 16
or 4y = 16
or y = 2
compareing set 1 and 4 we get R1/R4 = (0.01/0.07)z = 0.080/0.080 = 1
z= 0
Hence the rate law is R= k [A]1 [B]2 [C]0 = k [A]1 [B]2
b) From set 1 we can calculate the rate constant
Putting the values of R, [A] and [B] we get, 0.080 = k (0.15) (0.38) = k(0.057)
or, k = 0.080/(0.057) = 1.4035 L2 Mol-2 s-1
c) For the given stoichiometric reaction we can write
Rate (R) = -1/2 (dA/dt) = -(dB/dt) = -(dC/dt) = 1/3(dD/dt) = (dE/dt)
Hence the rate of appearence of D in set 3 is given by 0.0050x 3 = 0.015 M/s
ACS-1) the solution is supersaturated at 15 0C. (option b is the answer)
Reason: The definition of supersatuartion is that the solution contains more of the dissolved substance than it could be at the normal conditions.
ACS-2) option B is the answer
Calculation:
Molar mass of Nacl is 58.44
Hence 58.44 gms of Nacl dissolved in 1000 ml water will give 1 M Nacl solution
To prepare 250 ml 1 M solution we need (58.44/4) = 14.61 gm Nacl
To prepare 250 ml 0.10 M solution we need (14.61/10) = 1.461 gm Nacl
ACS-3) option A is the answer
Reason: Rate is given by dc/dt
For initial rate let us take dc= c2-c1= (0.4-0.3)= 0.1
dt= t2-t1= 5-0=5
hence, dc/dt = 0.1/5 = 0.02
ACS-4) The rate will be increased by a factor of 2. Option B is the answer
The rate law is R1= k [A] [B]2
Let the new rate is R2= k [A]/2 [2B]2 = 2 k [A] [B]2
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