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Q1

$\displaystyle rate = -\frac{d[H_2]}{dt} =-\frac{d[Br_2]}{dt} = \frac{1}{2} \frac{d[HBr]}{dt}$

Q2

$\displaystyle rate = - \frac{1}{2} \frac{d[NO_2]}{dt} = \frac{d[N_2O_4]}{dt}$

Q3

Coming soon

Q4

When

$t=t_{\frac{1}{2}}$ , $[X]=\frac{[X]_0}{2}$ .

$[X] =\frac{[X]_0}{2} = [X]_0e^{-kt_{\frac{1}{2}}}$

$\frac{1}{2} = e^{-kt_{\frac{1}{2}}}$

$t_{\frac{1}{2}} = \frac{\ln 2}{2k}$

$\ln 2$ is just a constant and as the rate constant does not vary with time, the half life is fixed.

Q5

In practice, the easiest way to verify a relationship is to choose variables which, when plotted against each other, give a straight line because it is much more straightforward to determine whether a line is straight compared to trying to confirm if a part of curve is polynomial or logarithmic, for example. It is also easier to measure parameters of a straight line (gradient, intercepts) from which other parameters can be calculated.

Given the expression:

$[X]=[X]_0e^{-kt}$

plotting [X] against t will yield a curve if the reaction is first order but it will hard to be sure the curve is exponential and also very difficult to deduce the value of k.

However, taking the natural logarithm of both sides yields

$\ln[X]=\ln[X]_0 -kt$

Now plotting ln[X] against t will yield a straight line (if the relationship is 1storder) with gradient equal to −k and vertical axis intercept equal to ln[X]₀.

Q6

Using the equation rate=k[X], a plot of initial rate against [X] will yield a straight line passing through the origin with gradient k.

A plot of rate against time is a little more complicated. We can write:

$\displaystyle rate=-\frac{d[X]}{dt} = -\frac{d}{dt}[X]_0e^{-kt}$

Thus we arrive at:

$rate=k[X]_0e^{-kt}$

Thus we see that the rate decays exponentially with just as [X] does. This makes sense since when [X] is high, its rate of loss will be high.

Q7

The definition of a half life yields:

$\displaystyle \frac{1}{2}[X]_0 = \frac{[X]_0}{kt_{\frac{1}{2}}[X]_0 +\ 1}$

$\displaystyle \frac{1}{2} = \frac{1}{kt_{\frac{1}{2}}[X]_0 +\ 1}$

$\displaystyle 2 = kt_{\frac{1}{2}}[X]_0 +\ 1$

$\displaystyle \Rightarrow t_{\frac{1}{2}} = {\frac{1}{k[X]_0}}$

At first glance this expression for half life also appears to be independent of time. Certainly kk is time-independent. However, it is important to remember that [X]₀ refers to the concentration of X at the start of the half life, not the very start of the reaction. The value of [X]₀ will be different for the first and second half lives and so on and is thus time dependent, making the half life itself time dependent.

Q8

$\displaystyle [X] = \frac{[X]_0}{kt[X]_0 +\ 1}$

$\displaystyle \frac{1}{[X]} = \frac{kt[X]_0 +\ 1}{[X]_0}$

$\displaystyle \Rightarrow \frac{1}{[X]} = kt +\ \frac{1}{[X]_0}$

A plot of $\displaystyle \frac{1}{[X]}$ against will be a straight line if the relationship is 2nd order. The gradient will be equal to k and the vertical axis intercept will be $\displaystyle \frac{1}{[X]_0}$ .

Q9

From rate=k[X]², plot of initial rate against [X] will be quadratic passing through the origin.

Q10

Q11

i. [X] to reach $\frac{1}{16}$ of its original value requires 4 half lives.

$\displaystyle t_{\frac{1}{2}} = \frac{\ln 2}{k} = 217 s$

$\displaystyle \Rightarrow 217 \time 4 = 868 s$

ii. [X] to reach $\frac{1}{e}$ of its original value - consider decay of [X]

$\displaystyle [X] = [X]_0 e^{-kt}$

[X] will reach $\frac{1}{e}$ of its original value when kt = 1

$\displaystyle t = \tau = \frac{1}{k}$

This is called the “lifetime” or “e-folding time” of X.

iii. Time taken for [Z] to reach $\frac{3}{4}$ of its final value is the same as the time taken for [X] to reach $\frac{1}{4}$ of its original value - i.e. 2 halves lives = 434 s.

Q12

$\displaystyle T = \frac{I}{I_0} = e^{-\epsilon c l}$

$\displaystyle \Rightarrow \epsilon = -\frac{1}{cl} \ln \frac{I}{I_0}$

$\displaystyle \frac{I}{I_0} = \frac{1}{7.14} = 0.14$

$\displaystyle \epsilon = -\frac{1}{1 \times 10^{-3} \times 2} \ln (0.14) = 983 mol^{-1} dm^{3} cm^{-1}$

Q13

$\displaystyle k = A e^{-\frac{E_a}{RT}}$

$\displaystyle \Rightarrow \ln k = \ln A - \frac{E_a}{RT}$

The pre-exponential factor can be found from the intercept of the vertical axis.

Q14

Consider one conversion at a time.

Moles → Molecules: multiply by Avagadro’s number N_A = 6.02 x 10²³

dm^-3 → cm^-3: Divide by 1000 as 1 dm³ is equivalent to 1000 cm³

s^-1 → min^-1: Multiply by 60

$\displaystyle 12 \times \frac{6.02 \times 10^{23} \times 60}{1000} = 4.3 \times 10^{23}$ molecules cm^-3 min^-1

Q15

i. $\displaystyle rate = k[A][B]$

$\displaystyle units = \frac{mol dm^{-3}s^{-1}}{mol^2 dm^{-6}=mol^{-1}dm^3s^{-1}}$

ii. $\displaystyle rate = k[A]^2[B]^{\frac{5}{2}}$

$\displaystyle units = \frac{mol dm^{-3}s^{-1}}{mol^{\frac{9}{2}} dm^{-\frac{21}{2}}} = mol^{-\frac{9}{2}}dm^{\frac{15}{2}}s^{-1}$

Q16

$\displaystyle E_{a-1} = E_{a1} - \Delta H$

Q17

i. $\displaystyle k_1 = A_1 e^{-\frac{E_{a1}}{RT}}$

ii. $\displaystyle k_{-1} = A_{-1} e^{-\frac{E_{a-1}}{RT}}$

Q18

$\displaystyle \frac{d[O]}{dt} = 2k_1[O_2] +\ k_2[O_3] - k_3[O][O_2][M] - k_4[O][O_3]$

Q19

The concentration of A can be calculated noting that the only reaction affecting A follows first order kinetics.

$\displaystyle \frac{d[A]}{dt} = k_1[A]$

$\displaystyle \Rightarrow [A]=[A]_0 e^{-k_1t}$

Thus [B] can be written as:

$\displaystyle \Rightarrow [B]=\frac{k_1}{k_2}[A] = \frac{k_1}{k_2} [A]_0 e^{-k_1t}$

Finally we can write

$\displaystyle \frac{d[C]}{dt} = k_2[B] = k_1[A]$

From this we note that the rise of [C] mirrors the decrease in [A], allowing us to write:

$\displaystyle -\Delta [A] = \Delta [C]$

$\displaystyle [C] = [A]_0 - [A]$

Thus the kinetics described on the “Concentration of Products” apply:

$\displaystyle [C] = [A]_0(1-e^{-k_1t})$

Q20

No we have:

$\displaystyle \frac{d[A]}{dt} = -k_1[A] +\ k_{-1}[B]$

$\displaystyle \frac{d[B]}{dt} = k_1[A] - k_{-1}[B] - k_{2}[B] +\ k_{-2}[C]$

$\displaystyle \frac{d[C]}{dt} = k_2[B] - k_{-2}[C]$

If B is in steady state, we can write:

$\displaystyle \frac{d[B]}{dt} = k_1[A] - k_{-1}[B] - k_{2}[B] +\ k_{-2}[C] = 0$

Rearranging:

$\displaystyle [B]_{ss} = \frac{k_1[A] +\ k_{-2}[C]}{k_{-1} +\ k_{2}}$