For an endothermic reaction, where ΔH represents the enthalpy of the reaction in kJ/mol, the minimum value for the energy of activation will be

1. Less than ΔH

2. Zero

3. More than ΔH

4. Equal to ΔΗ

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#3 | Factors Affecting Rate of Reaction

#4 | Arrhenius Equation

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Arrhenius Equation

(3) For endothermic reaction, energy of activation is always greater than $\u2206$H.

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How fast the reaction is at 25°C as compared to 0°C. If the activation energy is 65 KJ :-

(1) 2 times

(2) 5 times

(3) 11 times

(4) 16 times

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#3 | Factors Affecting Rate of Reaction

#4 | Arrhenius Equation

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Arrhenius Equation

(3) log k_{2}/k_{1} = E_{a}/2.303R(1/T_{1} - 1/T_{2})

Rate of the reaction at 0° C temperature =

Rate of the reaction at 25 °C temperature =

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The activation energy of a reaction is zero. The rate constant of the reaction is :

(1) increases with increase in temperature

(2) decreases with decrease in temperature

(3) decreases with increase in temperature

(4) is nearly independent of temperature

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#3 | Factors Affecting Rate of Reaction

#4 | Arrhenius Equation

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Arrhenius Equation

(4) K = Ae^{-Ea/RT}

K= Ae^{0/RT}

K = A= const.

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The rate constant of a first order reaction is 10^{-3} min^{-1} at 27°C. The temperature coefficient of this reaction is 2. The rate constant at 17°C will be :

(1) 10^{-3} min^{-1}

(2) 5 x 10^{-4} min^{-1}

(3) 2 x 10^{-3} min^{-1}

(4) 10^{-2} min^{-1}

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Arrhenius Equation

(2) K_{T1}/K_{T2} =$\mu $^{T/10}= 10^{-3}/K_{T1}=$\mu $^{10/10} = 2

K_{T1 }= 10^{-3}/2 = 0.5 x 10^{-3} =5x 10^{-4} min^{-1}

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Rate constant for a chemical reaction takes place at 500 K, is expressed as K = A e^{-1000}. The activation energy of the reaction is

(1) 100 cal/mol

(2) 1000 kcal/mol

(3) 10^{4} kcal/mol

(4) 10^{6} kcal/mol

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Arrhenius Equation

(2) K = A.e^{-Ea/RT}

Given, K =A.e^{-1000}

Ea/RT = 1000

Ea = 1000.2.500 = 1000000

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NEET - 2015

The activation energy of a reaction can be determined from the slope of which of the following graphs?

(a) In K vs T

(b) In K/T vs T

(c) In K vs 1/T

(d) T/In K vs 1/T

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#3 | Factors Affecting Rate of Reaction

#4 | Arrhenius Equation

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Arrhenius Equation

By arrhenius equation K = Ae^{-Ea/RT}

where, E_{a} = energy of activation

Applying log on both the side,

in K = inA - E_{a}/RT ...(i)

or log k= - E_{a}/2303RT + logA ....(ii)

This equation is of the form of y = mx+c i.e. the equation of a straight line. Thus, if a plot of log k vs 1/T is a straight line, the validity of the equation is confirmed.

Slope of the line = E_{a}/2.303R

Thus measuring the slope of the line, the value of E_{a }can be calculated .

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NEET - 2013

A reaction having equal energies of activation for forward and reverse reactions has

(a) $\u2206$S = 0

(b) $\u2206$G = 0

(c) $\u2206$H = 0

(d) $\u2206$H = $\u2206$G = $\u2206$S = 0

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#3 | Factors Affecting Rate of Reaction

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Arrhenius Equation

(c) Energy profile diagram for a reaction is as From the figure, it is clear that

${\left({\mathrm{E}}_{\mathrm{a}}\right)}_{\mathrm{b}}={\left({\mathrm{E}}_{\mathrm{a}}\right)}_{\mathrm{f}}+\u2206\mathrm{H}$

[Here (E_{a})_{b} = activation energy of backward reaction and (E_{a})_{f} activation energy of forward reaction].

If ${\left({\mathrm{E}}_{\mathrm{a}}\right)}_{\mathrm{b}}={\left({\mathrm{E}}_{\mathrm{a}}\right)}_{\mathrm{f}}$

then $\u2206\mathrm{H}=0$

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NEET - 2012

In a zero order reaction for every 10° rise of temperature, the rate is doubled. If the temperature is increased from 10°C to 100°C, the rate of the reaction will become

(a) 256 times

(b) 512 times

(c) 64 times

(d) 128 times

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Arrhenius Equation

(b) For 10° rise in temperature, n=1 so rate = 2^{n} = 2^{1} = 2

When temperature is increased from 10°C to 100°C, change in temperature = 100-10 = 90°C, i.e., n= 9

So, rate = 2^{9} 512 times

Alternate method With every 10° rise in temperature, rate becomes double, so

$\frac{r\text{'}}{r}={2}^{\left(\frac{100-10}{10}\right)}$ = 2^{9} = 512 times.

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NEET - 2010

For an endothermic reaction, energy of activation is E_{a} and enthalpy of reaction is $\u2206$H (both of these in kJ/mol). Minimum value of E_{a} will be

(a) less than $\u2206$H

(b) equal to $\u2206$H

(c) more than $\u2206$H

(d) equal to zero

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#3 | Factors Affecting Rate of Reaction

#4 | Arrhenius Equation

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Arrhenius Equation

Key Idea: In endothermic reactions, energy of reactants is less than that of the products. Potential energy diagram for endothermic reactions is,

where, E_{a} = activation energy of forward reaction

E'_{a} = activation energy of backward reaction

$\u2206$H = enthalpy of the reaction From the above diagram,

E_{a} = E'_{a} + $\u2206$H

Thus, Ea > $\u2206$H

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NEET - 2008

The rate constants k_{1} and k_{2} for two different reactions are 10^{16}e^{-2000/T} and 10^{15}e^{-1000/T } , respectively. The temperature at which k_{1}=k_{2} is:

(a) 1000 K

(b) $\frac{2000}{2.303}\mathrm{K}$

(c) 2000K

(d) $\frac{1000}{2.303}\mathrm{K}$

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#3 | Factors Affecting Rate of Reaction

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Arrhenius Equation

(d) Key Idea: The Arrhenius equation is represented as

$\mathrm{k}={\mathrm{Ae}}^{\raisebox{1ex}{$-{\mathrm{E}}_{\mathrm{a}}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{RT}$}\right.}$

In the given equations, first take log and then compare them.

${\mathrm{k}}_{1}={10}^{16}{\mathrm{e}}^{\raisebox{1ex}{$-2000$}\!\left/ \!\raisebox{-1ex}{$\mathrm{T}$}\right.}\phantom{\rule{0ex}{0ex}}{\mathrm{k}}_{2}={10}^{15}{\mathrm{e}}^{\raisebox{1ex}{$-1000$}\!\left/ \!\raisebox{-1ex}{$\mathrm{T}$}\right.}$

On taking log, we get

$\mathrm{log}{\mathrm{k}}_{1}=\mathrm{log}{10}^{16}-\frac{2000}{2.303\mathrm{T}}...\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\mathrm{log}{\mathrm{k}}_{2}=\mathrm{log}{10}^{15}-\frac{1000}{2.303\mathrm{T}}...\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}\because {\mathrm{k}}_{1}={\mathrm{k}}_{2}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{from}\mathrm{EQS}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}\mathrm{T}=\frac{1000}{2.303}\mathrm{k}$

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