# 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

C. Rate=1/2 delta C divided by delta temp
Explanation
The correct answer is rate=1/2 delta C divided by delta temp. This is because the reaction given in the question is A+3B--2C. The rate of the reaction is determined by the change in concentration of the reactants or products with respect to time. In this case, the rate is proportional to the change in concentration of C divided by the change in temperature. Therefore, the correct expression for the rate is rate=1/2 delta C divided by delta temp.

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• 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

C. 8
Explanation
When the concentration of B is doubled, the rate of reaction will increase by a factor of 8. This is because the rate law for the reaction is rate=k[A][B]3, which means that the rate is directly proportional to the concentration of B cubed. When the concentration of B is doubled, it will be raised to the power of 3, resulting in an increase of 2^3=8 in the rate of reaction.

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• 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

D. Decreasing the activation energy for the reaction
Explanation
A catalyst speeds up a reaction by decreasing the activation energy for the reaction. Activation energy is the minimum amount of energy required for a reaction to occur. By lowering this energy barrier, a catalyst allows the reaction to proceed more readily, increasing the rate of the reaction. The catalyst itself is not consumed in the reaction and can be used repeatedly.

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• 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

A. Zero-order
Explanation
The given graphs show a constant rate of reaction, regardless of the changes in concentration. This indicates that the reaction rate is independent of the concentration of the reactants. Therefore, this reaction is zero-order, meaning that the rate of reaction is constant and not influenced by the concentration of the reactants.

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• 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

C. Increasing the temperature
Explanation
Increasing the temperature would alter the rate constant (k) for the reaction. This is because temperature is directly proportional to the rate constant according to the Arrhenius equation. As temperature increases, the kinetic energy of the reactant molecules also increases, leading to more frequent and energetic collisions. This results in a higher rate of reaction and therefore a larger value of the rate constant.

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• 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

C. Increasing the temperature
Explanation
Increasing the temperature would alter the rate constant (k) for the reaction. This is because increasing the temperature increases the kinetic energy of the reactant molecules, leading to more frequent and energetic collisions. This results in an increase in the rate of the reaction, causing the rate constant (k) to increase.

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• 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

C. 17.4 kJ/mol
Explanation
The given equation represents a straight line in the form of y = mx + c, where y represents the natural logarithm of the rate constant (ln k), x represents the inverse temperature (1/T), and c represents the y-intercept. By comparing the equation with the given equation (y = -2.09 x 10^4*x + 343.3), we can see that the slope (m) is equal to -2.09 x 10^4. The activation energy (Ea) can be calculated using the formula Ea = -m * R, where R is the gas constant. Plugging in the values, we get Ea = -(-2.09 x 10^4) * 8.314 = 17.4 kJ/mol. Therefore, the activation energy is 17.4 kJ/mol.

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• 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]

A. Rate=k[NO2]2
Explanation
The given reaction is shown to be second order with respect to NO2, which means that the rate of the reaction is directly proportional to the square of the concentration of NO2. This is represented by the rate=k[NO2]2. The other options, rate=k[NO2][CO]2, rate=k[CO]2, and rate=k[NO][CO2], do not accurately represent the given information about the reaction order.

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• 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.

D. It double the proportion of molecules possessing at least the minimum energy required for the reaction.
Explanation
When the temperature increased from 10 degrees C to 20 degrees C, it caused the molecules to possess at least the minimum energy required for the reaction. This means that more molecules were able to overcome the activation energy barrier and participate in the reaction. As a result, the proportion of molecules possessing the minimum energy required for the reaction doubled.

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• 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]

A. K[A]2[B]2
Explanation
The rate of the reaction is directly proportional to the square of the concentration of A and the square of the concentration of B. This can be determined by comparing the initial rates of the reaction in the different experiments. In Experiment 1, doubling the concentration of A and B results in an increase in the initial rate by a factor of 4. In Experiment 2, doubling the concentration of A results in an increase in the initial rate by a factor of 4, while keeping the concentration of B constant. In Experiment 3, doubling the concentration of B results in an increase in the initial rate by a factor of 4, while keeping the concentration of A constant. Therefore, the rate law that most nearly accounts for these data is k[A]2[B]2.

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• 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

B. Z must react in a step after the rate determining step.
Explanation
The correct answer is "Z must react in a step after the rate determining step." This means that the formation of Z does not directly affect the rate of the reaction. The rate determining step is the slowest step in a reaction and determines the overall rate. Since Z is formed after this step, its concentration does not affect the rate.

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• 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

A. The forward reaction is exothermic
Explanation
The given correct answer states that the forward reaction is exothermic. This means that the forward reaction releases heat energy into the surroundings. In an exothermic reaction, the products have lower energy than the reactants, resulting in a negative change in enthalpy (Î”H). This can be inferred from the diagram provided, where the forward reaction is shown to release energy.

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• 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

B. 410s
• 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]

C. Kc=[H2][S2] divided by [H2S]2
Explanation
The equilibrium constant expression for a reaction is determined by the stoichiometry of the reaction. In this reaction, the stoichiometry is 2H2S(g) --> H2(g) + S2(g). This means that two molecules of H2S react to form one molecule each of H2 and S2. Therefore, the concentration of H2S is squared in the equilibrium constant expression to account for this stoichiometry. The concentrations of H2 and S2 are not squared because they appear as single molecules in the balanced equation. Therefore, the correct equilibrium constant expression is Kc=[H2][S2] divided by [H2S]2.

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• 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

C. 2
Explanation
The correct answer is 2. The tendency for a reaction to proceed to completion can be determined by the value of the equilibrium constant (Kp). The higher the value of Kp, the greater the tendency for the reaction to proceed to completion. In this case, the reactions are given with their respective Kp values. Comparing the Kp values, we can see that the reaction with the lowest Kp value (1.3 x 10 -5) has the least tendency to proceed to completion, while the reaction with the highest Kp value (5.9 x 10 -5) has the greatest tendency to proceed to completion. Therefore, the correct order of increasing tendency to proceed to completion is 2, 1, 3.

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• 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

A. K2=(K1)-2
Explanation
The answer K2=(K1)-2 is correct because it follows the relationship between the two equilibrium constants K1 and K2. According to the given equilibria, the forward reaction of the first equilibrium is the reverse reaction of the second equilibrium. Therefore, the equilibrium constant for the second equilibrium (K2) is the reciprocal of the equilibrium constant for the first equilibrium (K1) raised to the power of -2. This can be mathematically represented as K2=(K1)-2.

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• 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

D. [NO]2[Cl2]=Kc[NOCl]-2
Explanation
The answer [NO]2[Cl2]=Kc[NOCl]-2 is always true because it represents the equilibrium constant expression for the given reaction. The equilibrium constant (Kc) is defined as the ratio of the concentrations of the products (raised to their stoichiometric coefficients) to the concentrations of the reactants (raised to their stoichiometric coefficients), with each concentration term raised to the power of its coefficient in the balanced equation. In this case, the coefficient of NOCl is -2, so it is raised to the power of -2 in the equilibrium constant expression. Therefore, the given answer is the correct representation of the equilibrium constant expression for the reaction.

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• 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

C. No, right to left
Explanation
Based on the given information, the reaction is not at equilibrium. The equilibrium constant (Kc) is a large value, indicating that the reaction favors the formation of products (SO3). However, the concentrations of reactants (SO2 and O2) are higher than the concentration of the product (SO3), suggesting that the reaction has not yet reached equilibrium. Therefore, the reaction will occur in the direction from right to left in order to reach equilibrium and increase the concentration of the product.

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• 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.

• D.

Lower the temperature

B. Remove Br2
Explanation
Removing Br2 from the reaction will shift the equilibrium position to favor more products. This is because according to Le Chatelier's principle, when a reactant is removed from a system, the equilibrium will shift in the direction that produces more of that reactant. In this case, removing Br2 will cause the reaction to produce more NO and Br2, favoring the formation of products.

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• 20.

### Changing which of the following will change the equilibrium constant, K?

• A.

• B.

Doubling the volume

• C.

Decreasing the pressure

• D.

Increasing the temperature

D. Increasing the temperature
Explanation
Increasing the temperature will change the equilibrium constant, K, because it affects the balance between the reactants and products in a chemical reaction. According to Le Chatelier's principle, when the temperature is increased, the system will shift in the direction that absorbs heat, which is the endothermic reaction. This will cause the equilibrium position to shift towards the products or reactants, depending on whether the reaction is exothermic or endothermic. Consequently, the concentrations of the reactants and products will change, altering the value of the equilibrium constant, K.

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• 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.

C. Compress the gas mixture into a smaller volume
Explanation
Compressing the gas mixture into a smaller volume would cause the Br2 concentration to decrease. According to Le Chatelier's principle, when the volume of a gas is decreased, the system will try to counteract the change by shifting the equilibrium towards the side with fewer moles of gas. In this reaction, compressing the gas mixture would decrease the volume, causing the equilibrium to shift towards the side with fewer moles of gas, which is the side without Br2. As a result, the concentration of Br2 would decrease.

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• 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

C. The forward rate equals the reverse rate
Explanation
Chemical equilibrium is a state in a chemical reaction where the forward and reverse reactions occur at the same rate. This means that the rate at which reactants are converted into products is equal to the rate at which products are converted back into reactants. At this point, there is no net change in the concentrations of reactants and products, indicating that equilibrium has been reached.

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• 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)

A. K=0.116-1/s=In(0.88-In(1.76)
Explanation
The rate constant for the first order reaction A--B can be determined using the equation K = ln([A]t/[A]0) / t, where [A]t is the concentration of A at time t, [A]0 is the initial concentration of A, and t is the time elapsed. Using the given data, we can calculate the rate constant as follows: K = ln(0.88/1.76) / 6 = ln(0.5) / 6 = -0.693 / 6 = -0.116 s^-1. Therefore, the rate constant for the first order reaction A--B is -0.116 s^-1.

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• 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

A. K2=0.104
Explanation
The rate constant of a reaction is dependent on the activation energy (Ea) and the temperature. According to the Arrhenius equation, the rate constant (k) is proportional to the exponential of (-Ea/RT), where R is the gas constant and T is the temperature in Kelvin.

In this case, we are given the rate constant (k1) at 15 degrees C and we need to find the rate constant (k2) at 37 degrees C. Since the activation energy (Ea) is constant, we can use the Arrhenius equation to find the ratio of k2/k1.

By plugging in the values of Ea, R, T1 (15 degrees C + 273 = 288 K), and T2 (37 degrees C + 273 = 310 K), we can solve for the ratio of k2/k1.

The ratio k2/k1 is equal to 0.104, which means the rate constant at 37 degrees C (k2) is 0.104 times the rate constant at 15 degrees C (k1). Therefore, the correct answer is K2=0.104.

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• 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

A. Kc=[N2O4] divided by [NO2]2 =0.075/0.12=7.5
Explanation
The equilibrium constant, Kc, is calculated by dividing the concentration of N2O4 by the square of the concentration of NO2. In this case, the concentration of N2O4 at equilibrium is given as 0.0750 M. Therefore, to find Kc, we divide 0.0750 by the square of the concentration of NO2. The concentration of NO2 is not explicitly given in the question, but it can be determined by using the stoichiometry of the balanced equation. Since 2 moles of NO2 are produced for every mole of N2O4 that decomposes, the initial moles of NO2 can be calculated as (0.625 mol of N2O4) x (2 mol of NO2 / 1 mol of N2O4) = 1.25 mol of NO2. The initial concentration of NO2 can then be calculated as (1.25 mol of NO2) / (5.00 L of vessel) = 0.25 M. Taking the square of this concentration and dividing it into 0.0750 gives us a Kc value of 7.5.

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• 26.

### Hydrogen iodide decomposes according to the equation: 2HI(g)--H2(g)+I2 Kc=0.0156 at 400 degrees C If 0.550 mol HI was injected into a 2.0 L reaction vessel at 400 degrees C. Calculate the concentrtion of HI at equilibrium.

• A.

HI=0.275-2(0.02748)/square of [HI]=0.22M

• B.

HI=0.175 (0.175)/square of [HI]=0.33

A. HI=0.275-2(0.02748)/square of [HI]=0.22M
Explanation
The given answer correctly calculates the concentration of HI at equilibrium using the equation provided. It substitutes the values into the equation [HI] = 0.275 - 2(0.02748) / (0.275)^2, which simplifies to [HI] = 0.22 M. This value represents the concentration of HI at equilibrium in the reaction vessel.

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