Question

Describe the effect of concentration and temperature on the rate of chemical reaction. Rate constant K = 5.5 × 10⁻¹⁴ s⁻¹ for a first order reaction. Calculate the half life time of this reaction.

Hint:

More concentration and higher temperature give more effective collisions, so faster rate. For first order, \(t_{1/2} = 0.693/k\).

Answer 1

Step 1: Effect of concentration.
The rate of a reaction depends on the concentration of the reactants through the rate law, e.g. \(\text{Rate} = k[A]^x[B]^y\). As the concentration of a reactant increases, the number of reactant molecules per unit volume increases, so the frequency of effective collisions increases and the rate of reaction increases. As the reaction proceeds, reactants are used up, their concentration falls, and the rate decreases with time.

Step 2: Effect of temperature.
A rise in temperature increases the kinetic energy of the molecules, so a larger fraction of molecules have energy equal to or greater than the activation energy. This increases the number of effective collisions and hence the rate. Generally, the rate roughly doubles for every 10 K rise in temperature. Quantitatively, this is described by the Arrhenius equation \(k = A e^{-E_a/RT}\); as \(T\) increases, \(k\) increases.

Step 3: Half-life of a first order reaction.
For a first order reaction the half-life is independent of initial concentration and is given by:
\[ t_{1/2} = \frac{0.693}{k} \]
Step 4: Substitution.
\[ t_{1/2} = \frac{0.693}{5.5 \times 10^{-14}} \]
Step 5: Arithmetic.
\[ t_{1/2} = 0.126 \times 10^{14} = 1.26 \times 10^{13}\ \text{s} \]
\[\boxed{t_{1/2} = 1.26 \times 10^{13}\ \text{s}}\]