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Reactivity 2. How much, how fast and how far? - How fast? The rate of chemical change

Grade 11IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The rate of a chemical reaction is defined as the change in the concentration of a reactant or product per unit time. The standard units are mol dm−3 s−1mol \, dm^{-3} \, s^{-1}.

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Collision Theory: For a reaction to occur, particles must collide with an energy equal to or greater than the activation energy (EaE_{a}) and with the correct mutual orientation (geometry).

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Factors affecting the rate of reaction include: concentration of reactants, pressure (for gaseous reactions), surface area (for solids), temperature, and the presence of a catalyst.

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Increasing the temperature increases the rate of reaction because it increases the average kinetic energy of the particles. This results in a higher frequency of collisions and, more importantly, a much larger fraction of particles possessing energy E≥EaE \ge E_{a}.

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A catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy (EaE_{a}). It remains chemically unchanged at the end of the reaction.

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The Maxwell-Boltzmann distribution curve shows the distribution of kinetic energies in a sample of gas. As temperature increases, the peak of the curve shifts to the right (higher energy) and flattens, increasing the area under the curve to the right of EaE_{a}.

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Experimental methods to measure rates include: change in volume of gas evolved, change in mass (if a gas escapes), change in light transmission (colorimetry/spectrophotometry), and change in pHpH or conductivity.

📐Formulae

Average Rate=Δ[P]ΔtAverage \, Rate = \frac{\Delta [P]}{\Delta t}

Average Rate=−Δ[R]ΔtAverage \, Rate = -\frac{\Delta [R]}{\Delta t}

Rate (at time t)=gradient of tangent to the curve at time tRate \, (at \, time \, t) = \text{gradient of tangent to the curve at time } t

💡Examples

Problem 1:

In the decomposition of hydrogen peroxide, 2H2O2(aq)→2H2O(l)+O2(g)2H_{2}O_{2}(aq) \rightarrow 2H_{2}O(l) + O_{2}(g), the concentration of H2O2H_{2}O_{2} drops from 0.500 mol dm−30.500 \, mol \, dm^{-3} to 0.350 mol dm−30.350 \, mol \, dm^{-3} in 150 seconds150 \, seconds. Calculate the average rate of reaction in mol dm−3 s−1mol \, dm^{-3} \, s^{-1}.

Solution:

Rate=−Δ[H2O2]ΔtRate = -\frac{\Delta [H_{2}O_{2}]}{\Delta t} Rate=−0.350−0.500150Rate = -\frac{0.350 - 0.500}{150} Rate=−−0.150150Rate = -\frac{-0.150}{150} Rate=0.00100 mol dm−3 s−1Rate = 0.00100 \, mol \, dm^{-3} \, s^{-1}

Explanation:

The rate is calculated by dividing the change in concentration by the time interval. Because it is a reactant, the change is negative, so a negative sign is applied to ensure the rate value is positive.

Problem 2:

Explain why a 10 K10 \, K increase in temperature can lead to a doubling of the reaction rate even though the collision frequency only increases by about 2%2\%.

Solution:

While the total number of collisions increases slightly due to increased particle speed, the main effect of increasing temperature is seen in the Maxwell-Boltzmann distribution. A 10 K10 \, K rise significantly increases the proportion of molecules with kinetic energy E≥EaE \ge E_{a}.

Explanation:

The rate of reaction is more sensitive to the number of 'successful' collisions (those with sufficient energy) than to the total number of collisions. On the energy distribution curve, the area representing particles with E>EaE > E_{a} increases exponentially with temperature.