Hydrogen iodide, HI, decomposes in the gas phase to produce hydrogen, H2, and iodine, I2. The value of the rate constant, k, for the reaction was measured at several different temperatures, and the data are shown here:

Temperature(K)

k(M-1s-1)

555

6.23*10-7

575

2.42*10-6

645

1.44*10-4

700

2.01*10-3

What is the value of the activation energy (in kJ/mol) for this reaction?

Short Answer

Expert verified

The energy of activation for the reaction is 178.79 kJ/mol.

Step by step solution

01

Plot of ln K vs 1/T

T (K)

1/T (K-1)

k (M-1s-1)

ln k

555

0.0018

6.23*10-7

-14.2887

575

0.00174

2.42*10-6

-12.9317

645

0.00155

1.44*10-4

-8.84570

700

0.00143

2.01*10-3

-6.20962

The plot of ln k vs 1/T gives a straight line.

02

Calculation of Activation energy

The activated energy of a reaction is the minimum energy that a reactant must possess to convert into products.

The Arrhenius equation is given as

\(\begin{align}k &= A{e^{\frac{{ - {E_a}}}{{RT}}}}\\\ln k &= \ln A + (\frac{{ - {E_a}}}{{RT}})\end{align}\)

Here k is the rate constant; A is the pre-exponential factor, Eais the activation energy, R is the universal gas constant, and T is the temperature.

The value of\({{\bf{E}}_{\bf{a}}}\)can be calculated from the slope of the plot.

\({\bf{Slope = }}\frac{{{\bf{ - }}{{\bf{E}}_{\bf{a}}}}}{{\bf{R}}}\)

Considering the second and third data points on the plot, the slope is

\(\begin{align}Slope &= \frac{{\Delta y}}{{\Delta x}}\\\frac{{ - {E_a}}}{R} &= \frac{{ - 12.9317 - \left( { - 8.8457} \right)}}{{0.00174 - 0.00155}}\\ - {E_a} &= - 21505.26 \times 8.314\\{E_a} &= 178794.7J/mol\\{E_a} &= 178.79kJ/mol\end{align}\)

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Most popular questions from this chapter

Thefollowingdatahave been determined for the reaction: \({{\bf{I}}^{\bf{ - }}}{\bf{ + OC}}{{\bf{l}}^{\bf{ - }}} \to {\bf{I}}{{\bf{O}}^{\bf{ - }}} + {\bf{C}}{{\bf{l}}^{\bf{ - }}}\)

1

2

3

\({{\bf{(}}{{\bf{I}}^{\bf{ - }}}{\bf{)}}_{{\bf{initial}}}}\)(M)

0.10

0.20

0.30

\({{\bf{(OC}}{{\bf{l}}^{\bf{ - }}}{\bf{)}}_{{\bf{initial}}}}\)(M)

0.050

0.050

0.010

Rate(mol/l/s)

\({\bf{3}}{\bf{.5*1}}{{\bf{0}}^{{\bf{ - 4}}}}\)

\({\bf{6}}.{\bf{2*1}}{{\bf{0}}^{{\bf{ - 4}}}}\)

\({\bf{1}}.{\bf{83*1}}{{\bf{0}}^{{\bf{ - 4}}}}\)

Determine the rate equation and the rate constant for this reaction.

From the given data, use a graphical method to determine the order and rate constant of the following reaction: 2X⟶Y + Z

Time(s)

5.0

10.0

15.0

20.0

25.0

30.0

35.0.

40.0

(X)(M)

0.0990

0.0497

0.0332

0.0249

0.0200

0.0166

0.0143

0.0125

For the past 10 years, the unsaturated hydrocarbon 1,3-butadiene \(\left( {{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{ = CH - CH = C}}{{\bf{H}}_{\bf{2}}}} \right)\) has ranked 38th among the top 50 industrial chemicals. It is used primarily for the manufacture of synthetic rubber. An isomer exists also as cyclobutene:

The isomerization of cyclobutene to butadiene is first-order, and the rate constant has been measured as \({\bf{2}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{{\bf{s}}^{{\bf{ - 1}}}}\) at 150 \({\bf{^\circ C}}\) in a 0.53-L flask. Determine the partial pressure of cyclobutene and its concentration after 30.0 minutes if an isomerization reaction is carried out at 150 \({\bf{^\circ C}}\) with an initial pressure of 55 torr.

The rate of a certain reaction doubles for every 10°C rise in temperature.

(a) How much faster does the reaction proceed at 45°C than at 25°C?

(b) How much faster does the reaction proceed at 95°C than at 25°C?

If the initial concentration of butadiene is 0.0200 M, what is the concentration remaining after 20.0 min?

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