The rate of a certain reaction was studied at various temperatures. The table shows temperature (?)(T) and rate constant (?)(k) data collected during the experiments. Plot the data to answer the questions. What is the value of the activation energy, ?aEa , for this reaction?
The rate of a certain reaction was studied at various temperatures. The table shows temperature (?)(T) and rate constant (?)(k) data collected during the experiments. Plot the data to answer the questions.
What is the value of the activation energy, ?aEa , for this reaction?
| ? (?)T (K) | ? (?−1)k (s−1) |
|---|---|
| 400400 | 0.0001080.000108 |
| 420420 | 0.0006260.000626 |
| 440440 | 0.003100.00310 |
| 460460 | 0.01340.0134 |
| 480480 | 0.05110.0511 |
| 500500 | 0.1750.175 |
| 520520 | 0.5470.547 |
| 540540 | 1.571.57 |
| 560560 | 4.184.18 |
| 580580 | 10.410.4 |
Expert Answer
Given-
Table for temperature and rate constant values
| T (K) | k (s−1) |
|---|---|
| 400 | 0.000108 |
| 420 | 0.000626 |
| 440 | 0.00310 |
| 460 | 0.0134 |
| 480 | 0.0511 |
| 500 | 0.175 |
| 520 | 0.547 |
| 540 | 1.57 |
| 560 | 4.18 |
| 580 | 10.4 |
To find-
- Value of activation energy - Ea
- Value of pre‑exponential factor A
Two Forms of Arrhenius equation-
K= A e−EaRTe-EaRT
lnK = lnA - EaRTEaRT
were,
K = rate constant
A = pre‑exponential factor
Ea = activation energy
R = gas constant
T = temperature in K
Approach-
lnK = lnA - EaRTEaRT
This equation is in the form of-
y= mx + C
therefore,
y = lnk
x = 1T1T
C = lnA is intercept
m =- EaREaR is slope
- Graph can be plot by taking lnK values on x axis and 1T1T value on Y axis.
- slope = -EaREaR ; value of Ea can be found out form slope
- Calculating values of lnK and 1T1T
| T (K) | 1T1T | k (s−1) | lnK |
|---|---|---|---|
| 400 | 0.0025 | 0.000108 | -9.133 |
| 420 | 0.00238 | 0.000626 | -7.376 |
| 440 | 0.00227 | 0.00310 | -5.776 |
| 460 | 0.00217 | 0.0134 | -4.312 |
| 480 | 0.00208 | 0.0511 | -2.974 |
| 500 | 0.002 | 0.175 | -1.743 |
| 520 | 0.00192 | 0.547 | -0.603 |
| 540 | 0.00185 | 1.57 | -1.852 |
| 560 | 0.00179 | 4.18 | 1.43 |
| 580 | 0.00172 | 10.4 | 2.342 |
Equation form graph is
y= -13969 x + 25.889
Comparing with y= mx + c
therefore,
slope = m = -13969 =- EaREaR
C = 25.889 = lnA
-13969 =- Ea8.314 J mol−1 Ea8.314 J mol-1
Ea = 116138 J mol-1
Ea = 116138 Jmol Jmol x 1 KJ1000 J1 KJ1000 J = 116 KJmol KJmol
C = 25.889 = lnA
A = anti ln 25.889
A = 1.75 x 1011 s-1
Answer-
- Value of activation energy (Ea) = 116 KJmol KJmol
- Value of pre‑exponential factor A = 1.75 x 1011 s-1
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