WJEC A-Level Chemistry 1410QS
Formulae and Equations
Basic Ideas About Atoms
Chemical Calculations
Bonding
Solid Structures
The Periodic Table
Simple Equilibria and Acid-Base Reactions
Thermochemistry
Rates of Reaction
The Wider Impact of Chemistry
Organic Compounds
Hydrocarbons
Halogenoalkanes
Alcohols and Carboxylic Acids
Instrumental Analysis
Redox and Standard Electrode Potential
Redox Reactions
Chemistry of the P-Block
Chemistry of the D-Block Transition Metals
Chemical Kinetics
Enthalpy Changes for Solids and Solutions
Entropy and Feasibility of Reactions
Equilibrium Constants
Acid-Base Equilibria
Stereoisomerism
Aromaticity
Alcohols and Phenols
Aldehydes and Ketones
Carboxylic Acids and Their Derivatives
Amines
Amino Acids, Peptides and Proteins
Organic Synthesis and Analysis
Practical Skills and Techniques
Planning, Analysis and Evaluation
Core Experiments and Their Chemistry
Mathematical Skills in Chemistry
Mathematical Skills in Chemistry
Symbols, rearranging equations and logarithms: logarithms, the Arrhenius equation and rearranging formulae
- 2 things to know
- All units
Overview
pKw at 298 K, the relationship between ln and log₁₀, solving the Arrhenius equation for temperature and activation energy, half-life calculations, and rearranging mole and percentage yield formulae.
What you need to know
Answer each one before you reveal it.
A gaseous equilibrium has Kc = 4.0 × 10⁻³ mol dm⁻³. Comment on the position of equilibrium and express the comparison using an appropriate symbol.
Questions from this subtopic
Each one with the model answer underneath.
- 1
Show that pKw = 14.00 at 298 K and use it to relate pH and pOH.
Model answer
Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶, so pKw = −log₁₀(1.00 × 10⁻¹⁴) = 14.00. Taking negative logarithms of Kw = [H⁺][OH⁻] turns the product into a sum, giving pH + pOH = 14.00 at this temperature.
- 2
Explain why the Arrhenius equation is used in its logarithmic form when analysing rate data.
Model answer
Taking natural logarithms of k = Ae^(−Ea/RT) gives ln k = ln A − Ea/(RT). This has the form of a straight line, y = mx + c, with ln k plotted against 1/T. Rate constants vary over many orders of magnitude, so the logarithm both linearises the relationship and makes the values manageable.
- 3
How are ln x and log₁₀ x related, and why does the distinction matter in Arrhenius work?
Model answer
ln x = 2.303 log₁₀ x. It matters because the Arrhenius equation in logarithmic form uses natural logarithms; if base-ten logarithms are used by mistake, the gradient of the plot, and hence the activation energy, comes out wrong by a factor of 2.303.
- 4
A rate constant increases from 1.0 × 10⁻⁵ to 1.0 × 10⁻² s⁻¹ on heating. Express this change on a logarithmic scale and comment on why logarithms are appropriate here.
Model answer
ln(1.0 × 10⁻²) − ln(1.0 × 10⁻⁵) = ln(10³) = 6.91, and on a base-ten scale the change is exactly three units. Logarithms are appropriate because the rate constant has risen by a factor of one thousand, a change that a linear axis would render almost unreadable.
- 5
Why can taking a logarithm of a quantity with units be misleading, and how is this handled for pH and for ln k?
Model answer
Strictly, only a pure number can be the argument of a logarithm. In practice the quantity is understood to be divided by its standard unit, so pH uses the numerical value of [H⁺] in mol dm⁻³ and an Arrhenius plot uses the numerical value of k in its own units. This is why the units of k must be stated separately and are not carried into ln k.
- 6
A student takes logarithms of both sides of rate = k[A]ⁿ. What linear relationship results, and what is the gradient?
Model answer
Taking base-ten logarithms gives log₁₀(rate) = log₁₀ k + n log₁₀[A]. Plotting log₁₀(rate) against log₁₀[A] gives a straight line whose gradient is the order n and whose intercept is log₁₀ k.
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