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skelet666 [1.2K]
3 years ago
6

in a certain solvent is first order with respect to (CH3)3CBr and zero order with respect to OH2. In several experiments the rat

e constant k was determined at different temperatures. A plot of ln(k) versus 1/T was constructed that resulted in a straight line with a slope of 21.10 3 104 K and a y intercept of 33.5. Assume that k has units of s21. a. Determine the activation energy for this reaction. b. Determine the value of the frequency factor A. c. Calculate the value of k at 258C.
Chemistry
1 answer:
Inga [223]3 years ago
8 0

Answer:

Ea= -175.45J

A= 3.5×10^14

k=3.64 ×10^14 s^2.

Explanation:

From

ln k= -(Ea/R) (1/T) + ln A

This is similar to the equation of a straight line:

y= mx + c

Where m= -(Ea/R)

c= ln A

y= ln k

a)

Therefore

21.10 3 104= -(Ea/8.314)

Ea=-( 21.10 3 104×8.314)

Ea= -175.45J

b) ln A= 33.5

A= e^33.5

A= 3.5×10^14

c)

k= Ae^-Ea/RT

k= 3.5×10^14 × e^ -(-175.45/8.314×531)

k = 3.64 ×10^14 s^2.

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This is an exercise in the general or combined gas law.

To start solving this exercise, we must obtain the following data:

<h3>Data:</h3>
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We use the following formula:

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We clear the general formula for the final pressure.

\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{P_{1}V_{1}T_{2} }{V_{2}T_{1}}  \ \to \ Clear \ formula \end{gathered}$}

We solve by substituting our data in the formula:

\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{(6.54 \ atm)(4.5 \not{l})(367 \not{K}) }{(2.3 \not{l})(306 \not{k})}  \end{gathered}$}

\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{10800.81}{ 703.8 } \ atm  \end{gathered}$}

\boxed{\large\displaystyle\text{$\begin{gathered}\sf P_{2}=15.346 \ atm \end{gathered}$} }

If I raise the temperature to 94°C and decrease the volume to 2.3 liters, the pressure of the gas will be 15,346 atm.

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