Arrhenius Equation
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A graph is plotted for a real gas which follows Van der Waals equation with PVm taken on y-axis and P on x-axis. Find the intercept of the line where Vm is the molar volume.
In Born-Haber cycle, the sum of the enthalpy changes around a cycle is
Constant
Zero
Positive integer
Negative integer
- ΔH=−ve
ΔS=+ve - ΔH=+ve
ΔS=−ve - ΔH=+ve
ΔS=+ve - ΔH=−ve
ΔS=−ve
The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be:
(R=8.314JK−1 mol−1 and log2=0.301)
(IIT-JEE-2013)48.6kJmol−1
58.5kJmol−1
- 60.5kJmol−1
- 53.6kJmol−1
A→B, k1=1015e−2000/T
C→D, k2=1014e−1000/T
The temperature at which k1=k2 is:
- 2.303×2×313×27340log3
- 2.303×2×313×27340log(1/3) cal
- 2.303×2×40273×313log3 cal
- None of these
- 18231 J
- 12831 J
- 18431 J
- 19231 J
Let’s assume I want to double the rate of a reaction. The constraint is that I can only increase the temperature by 10 K. Calculate the minimum activation energy required if the temperature is 300K.
54 kJmol−1
64 kJmol−1
94 kJmol−1
84 kJmol−1
- 9.76s
- 5.76s
- 8.76s
- 3.76s
- 18231 J
- 12831 J
- 18431 J
- 19231 J
For a first order reaction A ⟶ P, the temperature (T) dependent rate constant (k) was found to follow the equation log k = -2000(1/T) + 6.0. The pre-exponential factor A and the activation energy Ea, respectively, are
(IIT-JEE, 2009)1.0×106s−1 and 9.2kJmol−1
6.0s−1 and 16.6kJ mol−1
1.0×106 s−1 and 16.6kJ mol−1
- 1.0×106 s−1 and 38.3kJ mol−1
(log2=0.3, log3=0.5, R=253 JK−1mol−1)
- 46 kJ mol−1
- 35 kJ mol−1
- 84 kJ mol−1
- 30 kJ mol−1
log10 k=20.35−(2.47×103)T
The energy of activation in kJ mol−1 is
(Nearest integer)
[Given : R=8.314 JK−1mol−1]
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