The rate of a homogeneous reaction is normally defined per unit
Answer: A
For a homogeneous reaction the volume is the natural basis, while catalytic rates are defined per unit mass or area of catalyst. Stating the basis is part of stating the rate at all. Concentration is the natural variable for the fluid phase.
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Assertion (A): The rate of a heterogeneous catalytic reaction is defined per unit mass of catalyst rather than per unit volume.
Reason (R): The reaction takes place on the catalyst surface rather than throughout the fluid.
Answer: B
Both are true and the reason gives the justification: doubling the fluid volume without adding catalyst does not double the rate, so volume is not the quantity the rate is proportional to. The catalyst is what the reaction actually needs.
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Reaction order differs from molecularity in that order is
Answer: C
Correct answer: determined experimentally and need not be a whole number
Order is an empirical description of how the rate responds to concentration, obtained by fitting data, and it may be fractional, zero or even negative - a chain reaction commonly shows an order of one half, and a product-inhibited reaction a negative one. Molecularity counts the molecules taking part in one elementary step and is necessarily a small positive integer.
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Consider the following statements about order and molecularity.
1. Order is determined from experimental rate data.
2. Molecularity applies only to an elementary step.
3. Order and molecularity are always numerically equal.
Answer: D
Correct answer: 1 and 2 only
Statements 1 and 2 are correct: order comes from experiment, and molecularity is meaningful only for an elementary step, since an overall equation is a summary rather than a mechanism. Statement 3 is true only for elementary reactions and false in general - the equality is the exception that identifies a step as elementary, not a rule that applies to overall reactions.
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State whether the following is true or false.
Reaction order can be fractional or negative.
Answer: A
A chain reaction may show an order of one and a half, and a species that inhibits a reaction appears with a negative exponent. Both are perfectly ordinary results from fitting a rate law. Chain reactions often show half integer orders.
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A reaction is second order overall. What are the units of its rate constant when the rate is in moles per litre per second?
Answer: B
Dividing the rate units by concentration squared leaves litres per mole per second. Checking the units of a rate constant is the quickest way to confirm the order it was fitted for. The exponent decides what units remain.
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Match each reaction order with the units of its rate constant, with rate in moles per litre per second.
List I
A. Zero order
B. First order
C. Second order
D. Third order
List II
1. litres per mole per second
2. moles per litre per second
3. litres squared per mole squared per second
4. per second
Answer: C
Correct answer: A-2, B-4, C-1, D-3
The units follow from \( \mathrm{rate} = kC^{n} \): since the rate is fixed in moles per litre per second, \( k \) must carry whatever units are needed to cancel \( C^{n} \). So \( k \) has the units of rate at zero order, reciprocal seconds at first, litres per mole per second at second, and litres squared per mole squared per second at third. The pattern is that each additional order adds a factor of litres per mole.
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Assertion (A): The units of a rate constant reveal the order of the reaction.
Reason (R): The constant must carry whatever units make the rate expression dimensionally consistent.
Answer: D
Both are true and the reason gives the mechanism: the concentration term supplies units raised to the order, so the constant must supply the remainder for the product to come out as a rate. A units check is quicker than refitting the data.
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The Arrhenius equation expresses the rate constant as
Answer: A
The exponential dependence is why modest temperature changes have such large effects on rate. Plotting the logarithm of the constant against reciprocal temperature gives a straight line of slope minus the activation energy over R.
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A reaction has an activation energy of about 50 kilojoules per mole. Roughly what happens to its rate for a ten degree rise near room temperature?
Answer: B
Correct answer: it roughly doubles
From the Arrhenius equation, \[ \frac{k_2}{k_1} = \exp\left[\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\right] = \exp\left[\frac{50{,}000}{8.314}\left(\frac{1}{298}-\frac{1}{308}\right)\right] = e^{0.655} = 1.9. \] The familiar rule that rate doubles for a ten degree rise is therefore not a law but a coincidence of the activation energies most ordinary reactions happen to have - a reaction of 25 kJ/mol would rise by only about forty per cent.
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