Cherryl's Chem Test #3
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Practice test #3 for chem II
- 1.For the hypothetical reaction A+3B--2C, the rate should be expressed as:
- A.
Rate=deltal A divided by deta temp
- B.
Rate=7/3 delta B divided by delta temp
- C.
Rate=1/2 delta C divided by delta temp
- D.
-1/2 delta C divided by delta temp
- 2.The reaction, A+2B--products; has the rate law, rate=k[A][B]3. When the concentration of B is doubled, while that of A is unchanged, by what factor will the rate of reaction increase?
- A.
2
- B.
4
- C.
8
- D.
6
- 3.A catalyst speeds up a reaction by
- A.
Increasing the number of high-energy molecules
- B.
Increasing the temperature of the molecules in the reaction
- C.
Increasing the number of collisions between molecules
- D.
Decreasing the activation energy for the reaction
- 4.The graphs below all refer to the same reaction. What order is this reaction?
- A.
Zero-order
- B.
First-order
- C.
Second-order
- D.
Third-order
- 5.Which one of the following would alter the rate constant (k) for the reaction below? 2A+B--products
- A.
Increasing the concentration of A
- B.
Increasing the concentration of B
- C.
Increasing the temperature
- D.
Measuring k again after the reaction has run a while
- 6.Which one of the following would alter the rate constant (k) for the reaction below? 2A+B--products
- A.
Increasing the concentration of A
- B.
Increasing the concentration of B
- C.
Increasing the temperature
- D.
Measuring k again after the reaction has run awhile
- 7.A plot of In k versus the inverse temperature (K-1) yielded a straight line with the equation below for a particular chemical reaction. If the gas constant (R) is 8.314 J/mol*K, calculate the activation energy in kJ/mole. y=-2.09 x 10 4th power *x +343.3
- A.
2.51 x 10 -5 kJ/mol
- B.
39.8 kJ/mol
- C.
17.4 kJ/mol
- D.
5.76 x 10 -6 kj/mol
- 8.I the reaction belwo was shown to be second order with respect to NO2, and zero order with respect to CO, then the rate law would be written as: NO2 (g) + CO (g)---NO(g)+C)23(g)
- A.
Rate=k[NO2]2
- B.
Rate=k[NO2][CO]2
- C.
Rate=k[CO]2
- D.
Rate=k[NO][CO2]
- 9.A change in temperature from 10 degrees C to 20 degree C is found to double the rate of a given chemical reaction. How did this change affect the reacting molecules?
- A.
It doubled their average velocity
- B.
It doubled their average energy
- C.
It doubled the number of collisions per second
- D.
It double the proportion of molecules possessing at least the minimum energy required for the reaction.
- 10.Consider a hypothetical reation 2A+B--products. Given the following information concerning the initial rate of the reaction with different initial concentrations: [A] (mol/L) [B] (mol/L) Initial Rate (M/s) Exp. 1 0.020 0.020 4.20 x 10 -3 Exp. 2 0.040 0.020 1.68 x 10 -2 Exp. 3 0.040 0.040 6.72 x 10 -2 What is the rate law that mostly nearly accounts for these data? Rate =
- A.
K[A]2[B]2
- B.
K[A][B]2
- C.
K[A][B]
- D.
K[A]2[B]
- 11.For the reaction: X2+Y=Z--XY+XZ The rate equation is: rate=k[X2][Y]. Why does the concentration of Z have no effect on the rate?
- A.
The concentration of Z is very small and the others are very large
- B.
Z must react in a step after the rate determining step.
- C.
Z is an intermediate
- D.
The activation energy for Z to react is very high
- 12.For the chemical reaction system described by the diagram below which statement is true?
- A.
The forward reaction is exothermic
- B.
The reverse reaction is exothermic
- C.
At equilibrium, the activation energy for the forward reaction is equal to the activation energy for the reverse reaction
- D.
The system is at equilibrium
- 13.The thermal decomposition of acetaldehyde is a second-order reaction. CH2CHO--CH4+CO What is the half-life of acetaldehyde,if the rate constant for the decomposition of acetaldehyde is 6.7 x 10 -6 mmHg*s?
- A.
1.5 x 10 5s
- B.
410s
- C.
5.4 x 10 7s
- D.
520s
- 14.The equilibrium constant expression for the reaction below is: 2H2S(g)--H2(g)+S2(g)
- A.
Kc=[H2S]2 divided by [H2][S2]
- B.
Kc=[H2S] divided by [H2][S2]
- C.
Kc=[H2][S2] divided by [H2S]2
- D.
Kc=[H2][S2] divided by [H2S]
- 15.For the following reactions occurring at 500K, arrange them in order of increasing tendency to proceed to completion(least to greatest tendency). 1. 2NOCl--2NO+Cl Kp=1.7 x 10 -2 2. 2SO3--2SO2+O2 Kp=1.3 x 10 -5 3. 2NO2--2NO+O2 Kp=5.9 x 10 -5
- A.
2
- B.
1
- C.
2
- D.
3
- 16.Consider the two gaseous equilibria: SO2 (g) + 1/20 (g)--SO3(g) K1 2SO3(g)--2SO2(g)+O2 (g) K2 The equilibrium constants K1 and K2 are related by:
- A.
K2=(K1)-2
- B.
K2 2nd power=K1
- C.
K2=(K1)2
- D.
K2=(K1)-1
- 17.When the following reaction is at equilibrium, which choice is always true? 2NOCl(g)--2NO(g)+Cl2(g)
- A.
[NO][Cl2]=[NOCl]
- B.
[NO]2[Cl2]=[NOCl]2
- C.
[NOCl]=[NO]
- D.
[NO]2[Cl2]=Kc[NOCl]-2
- 18.At 700K, the reaction: 2SO2(g)+O2--2SO3(g) has an equilibrium constant Kc=4.3 x 10 6th power, and the following concentrations are present: [SO2]=0.010M; [SO2]=10 M; [O2]=0.010M is the mixture at equilibrium, yes or not? if not at equilibrium, in which direction, left to right, or right to lef, will the reaction occur to reach equilibrium/
- A.
Yes
- B.
No, left to right
- C.
No, right to left
- D.
Ther is not enough information to tell
- 19.For the following reaction at equilibrium, which choice gives a change that will shift the position of equilibrium to favor more products? 2NOBR(g)---2NO(g)+Br2(g) delta Hrxn=+30 kJ
- A.
Increase the pressure
- B.
Remove Br2
- C.
Add more NO
- D.
Lower the temperature
- 20.Changing which of the following will change the equilibrium constant, K?
- A.
Adding a catalyst
- B.
Doubling the volume
- C.
Decreasing the pressure
- D.
Increasing the temperature
- 21.For the following reaction at equilibrium in a reaction vessel, which one of the changes below would cause the Br2 concentration to decrease? 2NOBr (g)--2NO(g)+Br2(g)
- A.
Remove some NO
- B.
Remove some NOBr
- C.
Compress the gas mixture into a smaller volume
- D.
Add some Br2
- 22.Chemical equilibrium has been reached when:
- A.
The forward rate exceeds the reverse rate
- B.
The reverse rate exceeds the forward rate
- C.
The forward rate equals the reverse rate
- D.
Concentration of the products are increasing
- 23.What is the rate constant for the first order reaction A--B given the following data: Time (s) [A] ---------------- 0 1.76 6 0.88 12 0.44 18 0.22
- A.
K=0.116-1/s=In(0.88-In(1.76)
- B.
K=1.16=In(0.77)-In(2.0)
- 24.Given that Ea for a certain biological reaction is 48kJ/mol, and that the rate constant is 2.5 x 10 -2 s-1 at 15 degrees C. What is the rate constant at 37 degrees C?
- A.
K2=0.104
- B.
K2=108.5
- C.
K2=220.0
- 25.The brown gas NO2 and the colorless gas N2O4 exist in equilibrium 2NO2--N2O4 In an experiment, 0.625 mol of N2O4 was introduced into a 5.00 L vessel and was allowed to decompose until it reached equilibrium with No2. The concentration of N2O4 at equilibrium was 0.0750 M. Calculate Kc for the reaction.
- A.
Kc=[N2O4] divided by [NO2]2 =0.075/0.12=7.5
- B.
Kc=[N2O4] divided by [NO2]2=0.065/0.11=6.5
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