Lesson test: Rate laws

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Rate laws

1.[2p]
Why can the orders of a rate law not be read off the balanced equation?

The orders describe which molecules meet in the slow step, which the overall equation does not show

Correct

Balanced equations are only approximate

The answer is: The orders describe which molecules meet in the slow step, which the overall equation does not show

Orders are always equal to the coefficients divided by two

The answer is: The orders describe which molecules meet in the slow step, which the overall equation does not show

The orders depend on the equilibrium constant

The answer is: The orders describe which molecules meet in the slow step, which the overall equation does not show

2.[2p]
A reaction is first order with $k = 4.87 \times 10^{- 3}$ s⁻¹. What is its half-life, in seconds?

CorrectNot quite: 142

3.[3p]
For the same reaction, how many seconds does it take for $90$ per cent of the reactant to be consumed?

CorrectNot quite: 473

4.[2p]
Doubling $[NO]$ quadruples the rate and doubling $[]$ doubles it. What is the overall order?

Correct

The answer is: 3

5.[1p]
The half-life of a second order reaction is independent of the starting concentration.

True

The answer is: False

False

Correct

6.[3p]
A reaction is second order with $k = 0.543$ dm³ mol⁻¹ s⁻¹ and starts at $0.0100$ mol dm⁻³. What is the concentration, in mol dm⁻³, after $100$ s?

CorrectNot quite: 0.00648

7.[3p]
Match each order to the plot that comes out straight.

Zero order
First order
Second order

log of concentration against time
reciprocal concentration against time
concentration against time

Show the answer

Zero order: concentration against time First order: log of concentration against time Second order: reciprocal concentration against time

8.[3p]
Which of these are reasons a reaction can appear zero order in a reactant?

Select all that apply

A catalyst surface is saturated

Correct

An enzyme is saturated with substrate

Correct

The rate is limited by the supply of light

Correct

The reactant is not involved in the reaction at all

The answer is: A catalyst surface is saturated, An enzyme is saturated with substrate, The rate is limited by the supply of light

9.[2p]
The rate law for $2 \to 3$ is $v = k []^{2} []^{- 1}$ . What does the negative order mean?

Oxygen inhibits the reaction, so the rate falls as product accumulates

Correct

Oxygen is consumed rather than produced

The answer is: Oxygen inhibits the reaction, so the rate falls as product accumulates

The reaction runs backward

The answer is: Oxygen inhibits the reaction, so the rate falls as product accumulates

The rate constant is negative

The answer is: Oxygen inhibits the reaction, so the rate falls as product accumulates

1.[2p] Why can the orders of a rate law not be read off the balanced equation?

2.[2p] A reaction is first order with k=4.87×10-3 s⁻¹. What is its half-life, in seconds?

3.[3p] For the same reaction, how many seconds does it take for 90 per cent of the reactant to be consumed?

4.[2p] Doubling [NO] quadruples the rate and doubling [] doubles it. What is the overall order?

5.[1p] The half-life of a second order reaction is independent of the starting concentration.

6.[3p] A reaction is second order with k=0.543 dm³ mol⁻¹ s⁻¹ and starts at 0.0100 mol dm⁻³. What is the concentration, in mol dm⁻³, after 100 s?

7.[3p] Match each order to the plot that comes out straight.

8.[3p] Which of these are reasons a reaction can appear zero order in a reactant?

9.[2p] The rate law for 2→3 is v=k[]2[]-1. What does the negative order mean?

1.[2p]
Why can the orders of a rate law not be read off the balanced equation?

2.[2p]
A reaction is first order with $k = 4.87 \times 10^{- 3}$ s⁻¹. What is its half-life, in seconds?

3.[3p]
For the same reaction, how many seconds does it take for $90$ per cent of the reactant to be consumed?

4.[2p]
Doubling $[NO]$ quadruples the rate and doubling $[]$ doubles it. What is the overall order?

5.[1p]
The half-life of a second order reaction is independent of the starting concentration.

6.[3p]
A reaction is second order with $k = 0.543$ dm³ mol⁻¹ s⁻¹ and starts at $0.0100$ mol dm⁻³. What is the concentration, in mol dm⁻³, after $100$ s?

7.[3p]
Match each order to the plot that comes out straight.

8.[3p]
Which of these are reasons a reaction can appear zero order in a reactant?

9.[2p]
The rate law for $2 \to 3$ is $v = k []^{2} []^{- 1}$ . What does the negative order mean?