9. Which one of the following changes would cause the number of collisions betwe
ID: 498146 • Letter: 9
Question
9. Which one of the following changes would cause the number of collisions between molecules of gas to DECREASE? all choices. (a) Decreasing the temperature of gas Decreasing the gas volume while keeping (c) ncreasing the gas volume while keeping its amount unchanged (d) Decreasing the molecular weight of gas (e) Increasing the molecular weight of gas 10. Initial rates shown below were measured for the following hypothetical reaction: What is value of the rate constant, k for this reaction? Trial Initial [A] Initial [B] nitial Rate (molL) mol/l min) 0.200 0.100 6.00 x 10-2 0.100 0.100 1.50 x 10-2 0.200 1.20 x 10-1 0.200 4 0.300 0.200 2.70 x 10-1 (a) 3.00 M min (b) 6.67x103 M min (c) 1.50x10 mini (d) 1.50x101 M min (e) 1.50x10' M 'min' 11. The gas-phase reaction CH3NC CHlCN has been studied in a closed vessel, and the rate equation was found to be: Rate -AICH3NCI/AM klCH3NC]. Which one of the following actions will increase the rate constant of the reaction? Mark all correct choices. Tom (a) lowering the temperature (b) adding a catalyst (c) using a larger initial amount of CH NC in the same vessel (d) using a bigger vessel, but the same initial amount of CHNC (e) increasing the temperature 12. A catalyst accelerates a reaction because (a) it increases the number of molecules with energy equal to or greater than the activation (b) it lowers the activation energy for the reaction. it increases the number of collisions between molecules. (c) (d) it increases the temperature of the molecules in the reaction. (e) it supplies energy to reactant molecules.Explanation / Answer
9. To decrease the number of collisions we have to decrease the kinetic energy of the particles and increase the length of the mean free path. In order to achieve these two parameters, we need to
(a) decrease the temperature to decrease the kinetic energy, and (C) increase the gas volume keeping the amount unchanged.
10. From trial 1 and 3, it is observed that as the initial concentration of B is doubled the initial rate of reaction also doubles keeping the initial concentration of A constant. Hence, it can be said that the rate of reaction is linearly dependent on the concentration of B.
From trial 1 and 2, it is observed that as the concentration of A is doubled, the rate of reaction is quadrupled. Hence, it can be said that the rate of reaction is dependent to the square of concentration of A.
The reaction rate can then be written as below:
-r, rate = k*[A]2*[B]
Therefore, k = rate/(A^2)*(B). Units of k are (mol/Lmin)/(mol/L)^3 or (M/min)/(M^3) = M-2min-1.
Therefore, from trial 1, k = (6*10^-2)/(0.2^2)*(0.1) = 1.5*101 M-2 min-1
11. According to Arrhenius law, the rate constant k = Ae-Ea/RT. Hence, the rate constant can be increased b decreasing the Activation energy (Ea) by adding a catalyst, or by increasing the temperature of the reaction.
12. A catalyst accelerates a reaction because it lowers the Activation energy for the reaction.
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