How many times the rate of reaction increases at 20^{0}C for a reaction having the activation energies in the presence and absence of catalyst as 50 kJ/mol and 75 kJ/mol?
K = Ae^(Ea/RT)
Taking ln on both sides
lnk = lnA + (Ea/R) x 1/T
Now ATQ: lnk1 = lnA + (Ea1/R) x 1/T
lnk2 = lnA + (Ea_{2}/R) x 1/T
Subtracting the two equations
ln(k_{2}/k_{1}) = (Ea_{1}  Ea_{2})/RT
ln(k_{2}/k_{1}) = (75  50) x 1000 / 8.314 x 293
K_{2}/K_{1} = e ^{(25000 / 8.314 x 293)}
K_{2}/K_{1} = e^{10.26}
This is approximately equal to 30000.
Hence C is correct.
If E_{f} and E_{b} are the activation energies of the forward and reverse reactions and the reaction is known to be exothermic, then:
The correct answer is Option B.
For exothermic reaction, ΔH<0
E_{b} = E_{f }+ ∣ΔH∣ or
E_{f} < E_{b}
The ratio of the rate constant of a reaction at two temperatures differing by __________0C is called temperature coefficient of reaction.
The ratio of the rate constant of a reaction at two temperatures differing by 10^{0}C is called temperature coefficient of reaction.
The effect of temperature on reaction rate is given by
The activation energies of two reactions are given as E_{a1}= 40 J and E_{a2}= 80 J, then the relation between their rate constants can be written as:
The correct answer is Option A.
As the value of activation energy, E_{a} increases, the value of rate constant, k decreases.
So, k_{1} > k_{2} since E_{1} < E_{2}
The activation energy of a chemical reaction can be determined by
The correct answer is Option C
Activation energy can be determined by evaluation rate constants at different temperatures using equation:
Where, K_{2 }= rate constant at temperature T_{2}
K_{1}= rate constant at temperature T_{1}
The rate constant, activation energy and Arrhenius parameter of a chemical reaction at 25°C are 3.0 x 10^{4} s^{1}, 104.4 kJ mol^{1} and 6.0 x 10^{14} s^{1 }respectively. The value of the rate constant at infinite temperature is is
The correct answer is option C
Arrhenius equation
⇒ K = Ae−Ea/RT
As T → ∞,
RT → ∞
e^{−Ea/RT} → 1
Hence K → A as T→∞
∴ Value of K as T → ∞ = 6.0×10^{14}S^{−1}
The plot of ln k vs 1/T is a straight line. The slope of the graph is:
The Arrhenius equation is k=Ae^{−Ea/RT}
ln k = ln A  E_{a}/RT
comparing with y = mx+c where y = ln k and x = 1/T, slope m becomes E_{a}/R
so when ln k vs 1/T is plotted, the slope comes out to be  E_{a}/R
For a chemical reaction the rate constant is nearly doubled with the rise in temperature by
B is the right answer.
Rate constant is doubled with 10 degree rise in temperature.
The reactions with low activation energy are
The correct answer is Option C.
The reactions with low activation energies are always fast whereas the reactions with high activation energy are always slow. The process of speeding up a reaction by reducing its activation energy is known as catalysis, and the factor that's added to lower the activation energy is called a catalyst.
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