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Mathematical Skills in Chemistry

Symbols, rearranging equations and logarithms: logarithms, the Arrhenius equation and rearranging formulae

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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.

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Key term

Rearrange the expression for percentage yield to find the theoretical mass of product, given a yield of 75.0% and an actual mass of 3.00 g.

Key term

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. 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. 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. 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. 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. 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. 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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