Past exam of the chemistry A course of the University of Cambridge / 2024 / Paper 2 / A7 / c / iii

Table of contents
- Solution iii

Solution

Using the numerical rate constants in

d m^{3} mo l^{- 1} s^{- 1}

, the logarithms are

p / atm ln k_{2} 100 - 7.58306 300 - 7.37936 700 - 6.96963 1000 - 6.66090

A straight-line least-squares fit is

ln k_{2} = - 7.68621 + (1.02471 \times 1 0^{- 3}) (p / atm) .

Every logarithmic residual has magnitude below

7.2 \times 1 0^{- 4}

, so the rates follow the fitted exponential to within

0.08%

;

R^{2} ≃ 0.999997

. Thus a constant activation volume describes these data well.

Figure 1.
Original logarithmic pressure-rate fit and residuals for the activation-volume calculation
. Details.

Convert the slope to pascals, using

1 atm = 101325

Pa and

T = 293.15

K:

Δ V^{‡\circ} = - R (293.15) \frac{1.02471 \times 1 0 ^{- 3}}{101325} m^{3} mo l^{- 1} = - 24.65 c m^{3} mo l^{- 1} .

Multiplying all rates by a common unit-conversion factor changes the intercept but leaves this slope and volume unchanged.

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