Kinetic Molecular Theory - AP Chemistry

Card 0 of 18

Question

Which of the following is not an assumption of the kinetic molecular theory of gases?

Answer

The kinetic molecular theory of gases states that the average kinetic energy of gas particles is proportional to temperature, and it is the same for all gases at a given temperature. This is the opposite of what is stated in the answer choice.

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Question

Two moles of nitrogen gas are kept in a glass container. The temperature of the gas is 400K.

What is the kinetic energy of the gas?

$R=8.314\frac{J}{mol*K}$

Answer

Using the equation $KE = \frac{3}{2}nRT$ , n being the number of moles, R being 8.314, and T being the temperature in Kelvin, we can find the kinetic energy of two moles of gas in a container.

$KE = \frac{3}{2}(2 mol)(8.314\frac{J}{mol*K})(400 K)$

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Question

Why does a higher temperature result in a faster reaction?

Answer

The temperature of a system can be manipulated to increase the reaction rate. The temperature affects the rate of the reaction because as temperature increases, so does the average kinetic energy. This means that molecules (reactants) are moving around more quickly, resulting in more frequent molecular collisions, resulting in an increased rate of product formation.

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Question

Which of the following statements in not an assumption made in kinetic molecular theory?

Answer

The average kinetic energy is considered to be a sole function of temperature; pressure is not a factor. All of the other statements are assumptions of kinetic molecular theory.

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Question

The reaction $A+B\rightarrow C$ obeys the rate law $r=-k[A]^{1/2}[B]^{2/3}$

What are the units of the rate constant, ?

Answer

Recall that $r[=]\frac{M}{s}$ and that and are both molar concentrations. Then,

$\frac{M}{s} = k(\frac{M}{s})^{1/2}(\frac{M}{s})^{2/3}=k(\frac{M}{s})^{7/6}$

Dividing both sides by $(\frac{M}{s})^{7/6}$ gives

$k[=]M^{-1/6}s^{-1}$

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Question

Consider the following first order reaction

$A\overset{k}{\rightarrow}B$

Where k is the rate constant.

$k=0.001s^{-1}$

and the initial concentrations of A and B are

$[A]_{0}=3 mols$

$[B]_{0}=0mols$

What is the percent conversion after 500 seconds? Round to the nearest percent.

Answer

For a first order reaction,

$r=\frac{dC_{A}}{dt}=-kC_{A}$

$\frac{dC_{A}}{C_{A}}=-kdt$

Integrating both sides gives

$ln(C_{A})=-kt+C$

Where is an integration constant.

Solving for gives

$C_{A}=C_{1}e^{-kt}$

Initially there are 3 mols of . Using this we find that

$C_{1}=3$

$C_{A}=3e^{-kt}$

At ,

$C_{A}=3e^{-0.001:s^{-1}500:s}=1.8196 :mols$

Percent conversion is calculated as follows:

$X_{A}=\frac{Initial-Final}{Initial}100 = \frac{3-1.8196}{3}100=39%$

Compare your answer with the correct one above

Question

Which of the following is not an assumption of the kinetic molecular theory of gases?

Answer

The kinetic molecular theory of gases states that the average kinetic energy of gas particles is proportional to temperature, and it is the same for all gases at a given temperature. This is the opposite of what is stated in the answer choice.

Compare your answer with the correct one above

Question

Two moles of nitrogen gas are kept in a glass container. The temperature of the gas is 400K.

What is the kinetic energy of the gas?

$R=8.314\frac{J}{mol*K}$

Answer

Using the equation $KE = \frac{3}{2}nRT$ , n being the number of moles, R being 8.314, and T being the temperature in Kelvin, we can find the kinetic energy of two moles of gas in a container.

$KE = \frac{3}{2}(2 mol)(8.314\frac{J}{mol*K})(400 K)$

Compare your answer with the correct one above

Question

Why does a higher temperature result in a faster reaction?

Answer

The temperature of a system can be manipulated to increase the reaction rate. The temperature affects the rate of the reaction because as temperature increases, so does the average kinetic energy. This means that molecules (reactants) are moving around more quickly, resulting in more frequent molecular collisions, resulting in an increased rate of product formation.

Compare your answer with the correct one above

Question

Which of the following statements in not an assumption made in kinetic molecular theory?

Answer

The average kinetic energy is considered to be a sole function of temperature; pressure is not a factor. All of the other statements are assumptions of kinetic molecular theory.

Compare your answer with the correct one above

Question

The reaction $A+B\rightarrow C$ obeys the rate law $r=-k[A]^{1/2}[B]^{2/3}$

What are the units of the rate constant, ?

Answer

Recall that $r[=]\frac{M}{s}$ and that and are both molar concentrations. Then,

$\frac{M}{s} = k(\frac{M}{s})^{1/2}(\frac{M}{s})^{2/3}=k(\frac{M}{s})^{7/6}$

Dividing both sides by $(\frac{M}{s})^{7/6}$ gives

$k[=]M^{-1/6}s^{-1}$

Compare your answer with the correct one above

Question

Consider the following first order reaction

$A\overset{k}{\rightarrow}B$

Where k is the rate constant.

$k=0.001s^{-1}$

and the initial concentrations of A and B are

$[A]_{0}=3 mols$

$[B]_{0}=0mols$

What is the percent conversion after 500 seconds? Round to the nearest percent.

Answer

For a first order reaction,

$r=\frac{dC_{A}}{dt}=-kC_{A}$

$\frac{dC_{A}}{C_{A}}=-kdt$

Integrating both sides gives

$ln(C_{A})=-kt+C$

Where is an integration constant.

Solving for gives

$C_{A}=C_{1}e^{-kt}$

Initially there are 3 mols of . Using this we find that

$C_{1}=3$

$C_{A}=3e^{-kt}$

At ,

$C_{A}=3e^{-0.001:s^{-1}500:s}=1.8196 :mols$

Percent conversion is calculated as follows:

$X_{A}=\frac{Initial-Final}{Initial}100 = \frac{3-1.8196}{3}100=39%$

Compare your answer with the correct one above

Question

Which of the following is not an assumption of the kinetic molecular theory of gases?

Answer

The kinetic molecular theory of gases states that the average kinetic energy of gas particles is proportional to temperature, and it is the same for all gases at a given temperature. This is the opposite of what is stated in the answer choice.

Compare your answer with the correct one above

Question

Two moles of nitrogen gas are kept in a glass container. The temperature of the gas is 400K.

What is the kinetic energy of the gas?

$R=8.314\frac{J}{mol*K}$

Answer

Using the equation $KE = \frac{3}{2}nRT$ , n being the number of moles, R being 8.314, and T being the temperature in Kelvin, we can find the kinetic energy of two moles of gas in a container.

$KE = \frac{3}{2}(2 mol)(8.314\frac{J}{mol*K})(400 K)$

Compare your answer with the correct one above

Question

Why does a higher temperature result in a faster reaction?

Answer

The temperature of a system can be manipulated to increase the reaction rate. The temperature affects the rate of the reaction because as temperature increases, so does the average kinetic energy. This means that molecules (reactants) are moving around more quickly, resulting in more frequent molecular collisions, resulting in an increased rate of product formation.

Compare your answer with the correct one above

Question

Which of the following statements in not an assumption made in kinetic molecular theory?

Answer

The average kinetic energy is considered to be a sole function of temperature; pressure is not a factor. All of the other statements are assumptions of kinetic molecular theory.

Compare your answer with the correct one above

Question

The reaction $A+B\rightarrow C$ obeys the rate law $r=-k[A]^{1/2}[B]^{2/3}$

What are the units of the rate constant, ?

Answer

Recall that $r[=]\frac{M}{s}$ and that and are both molar concentrations. Then,

$\frac{M}{s} = k(\frac{M}{s})^{1/2}(\frac{M}{s})^{2/3}=k(\frac{M}{s})^{7/6}$

Dividing both sides by $(\frac{M}{s})^{7/6}$ gives

$k[=]M^{-1/6}s^{-1}$

Compare your answer with the correct one above

Question

Consider the following first order reaction

$A\overset{k}{\rightarrow}B$

Where k is the rate constant.

$k=0.001s^{-1}$

and the initial concentrations of A and B are

$[A]_{0}=3 mols$

$[B]_{0}=0mols$

What is the percent conversion after 500 seconds? Round to the nearest percent.

Answer

For a first order reaction,

$r=\frac{dC_{A}}{dt}=-kC_{A}$

$\frac{dC_{A}}{C_{A}}=-kdt$

Integrating both sides gives

$ln(C_{A})=-kt+C$

Where is an integration constant.

Solving for gives

$C_{A}=C_{1}e^{-kt}$

Initially there are 3 mols of . Using this we find that

$C_{1}=3$

$C_{A}=3e^{-kt}$

At ,

$C_{A}=3e^{-0.001:s^{-1}500:s}=1.8196 :mols$

Percent conversion is calculated as follows:

$X_{A}=\frac{Initial-Final}{Initial}100 = \frac{3-1.8196}{3}100=39%$

Compare your answer with the correct one above

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