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Consider the reaction described by the following equation

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Consider the reaction described by the following equation. C2H4Br2(aq) + 3I-(aq) rightarrow C2H4(g) + 2Br-(aq) + (aq) The rate law is rate = k[C2H4Br2][I-] where k=5.33*10-3 M-1.s-1 What are the missing entries in the following table? Consider this initial-rate data at a certain temperature for the reaction described by C2H5Cl(g) rightarrow C2H4 + HCl(g) Determine the value and units of the rate constant. The following reaction has an activation energy of 262 kJ/mol. c4H8(g) rightarrow 2C2H4(g) At 600.0 K the rate constant is 6.1 Times 10-5 s-1. What is the value of the rate constant at 815.0 K? Consider this reaction data: If you were going to graphically determine the activation energy of this reaction, what points would you plot? To avoid rounding errors, use at least three significant figures in all values. Determine the rise, run, and slope of the line formed by these points. What is the activation energy of this reaction?

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Question 12.6

We are fortunately given the rate law, which will make this question much less tedious than it could be. Consider the data for run 1, which we can plug into the rate law as follows:

rate = k[C₂H₄Br][I^-]

0.00114 M s^-1 = (5.33 x 10^-3 M^-1 s^-1)[x][0.231 M]

Solving for x, we get x = 0.926 M

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Now we do the same thing to find y, we just need to use run 2's data:

0.000569 M s^-1 = (5.33 x 10^-3 M^-1 s^-1)[0.231 M][y]

This gives y = 0.462 M

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Again, use run 3's data to find z.

rate = (5.33 x 10^-3 M^-1 s^-1)[0.231 M][0.231 M] = 0.000284 M s^-1 = z

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Question 12.12

We must determine the order of reaction first. This can be done by using the relation:

$\frac{rate \ 2}{rate \ 1} = (\frac{[C_{2}H_{5}Cl]_{2}}{[C_{2}H_{5}Cl]_{1}})^{n}$

Let's use data from run's 1 and 2 as follows to solve for n, the order of reaction:

$\frac{1.12 \times 10^{-30} M/s}{0.560 \times 10^{-30} M/s} = (\frac{[0.200 M]}{[0.100 M]})^{n}$

$2 = 2^{n}$

Thus n = 1, so the reaction is first order with respect to C₂H₅Cl. Then the rate law is:

rate = k[C₂H₅Cl].

Now we can use the data from any run to determine k. Let's use the data from run #1:

0.560 x 10^-30 M/s = k(0.100 M)

k = 5.6 x 10^-30 s^-1

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Question 12.25

We can use the two-point Arrhenius equation to find the value of a rate constant:

$ln(\frac{k_{2}}{k_{1}}) = \frac{-E_{a}}{R}(\frac{1}{T_{2}}-\frac{1}{T_{1}})$

Let k₂ be the rate constant we are solving for. That means k₁ = 6.1 x 10^-8 s^-1 with T₁ = 600.0 K and T₂ = 815.0 K. So:

$ln(\frac{k_{2}}{6.1 \times 10^{-8} s^{-1}}) = \frac{-262 \times 10^{3} J/mol}{8.314 \ J/mol K}(\frac{1}{815 \ K}-\frac{1}{600.0 \ K})$

$ln(\frac{k_{2}}{6.1 \times 10^{-8} s^{-1}}) = 13.855$

$\frac{k_{2}}{6.1 \times 10^{-8} s^{-1}} = 1.04 \times 10^{6}$

k₂ = 0.0634 s^-1

This answer ought to make sense because the rate of reaction increases with temperature.

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Question 12.26

The Arrhenius equation can also take the form of

$k = Ae^{\frac{-E_{a}}{RT}}$

$ln(k) = \frac{-E_{a}}{RT}+ln(A)$

Which can be written as:

$ln(k) = \frac{-E_{a}}{R}(\frac{1}{T})+ln(A)$

Notice that this equation is in the form y = mx + b, where x = 1/T and m = -E_a/R.

So the points you need to use are ln(0.868) = -0.141563 and ln(0.380) = -0.96758 for your y-values. For your x values, you need to use 1/725 K = 0.0013793 K^-1 and 1/325 K = 0.0030769 K^-1. So the points are:

(0.0013793 K^-1, -0.141563) and (0.0030769 K^-1, -0.96758)

The rise is given by:

$\Delta y = y_{2}-y_{1} = -0.96758 - (-0.141563) = -0.826017$

The run is given by:

$\Delta x = x_{2}-x_{1} = 0.0030769 \ K^{-1} - 0.0013793 \ K^{-1} = 0.0016976 K^{-1}$

The slope is just the rise over the run, so m = -0.826017/0.0016976 K^-1 = -486.58 K

The activation energy form the Arrhenius equation above has the relation:

m = -E_a / R

-mR = E_a

E_a = -(-486.58 K)(8.314 J/mol K) = 4045.426 J/mol

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Consider the reaction described by the following equation

Chemistry