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Byju's Answer

Standard XII

Chemistry

Arrhenius Equation

Consider the ...

Question

Consider the following statements:
(i) Increase in concentration of reactant increases the rate of a zero order reaction.
(ii) Rate constant $k$ is equal to collision frequency $A$ if $E_{a} = 0$ .
(iii) Rate constant $k$ is equal to collision frequency $A$ if $E_{a} = \infty$ .
(iv) ${log}_{e} k$ vs $T$ is a straight line.
(v) ${log}_{e} k$ vs $1 / T$ is a straight line.
Correct statements are:

(i) and (iv)

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(ii) and (v)

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(iii) and (iv)

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(ii) and (iii)

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Solution

The correct option is B (ii) and (v)
$k = A e^{- E_{a} / k_{B} T}$
So if $E_{a} = 0,$ then $k = A .$ also,
${log}_{e} k = {log}_{e} (A) - \frac{E_{a}}{k_{B} T}$
So, ${log}_{e} k$ us $\frac{1}{T}$ is a straight line.

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Similar questions

According to Arrhenius equation rate constant k is equal to $A . e^{- E_{a} / R T .}$

Which of the following options represents the graph of ln k vs $\frac{1}{T}$ ?

Q. Graph between log K and

\frac{1}{T}

[Where K is rate constant

(S^{- 1})

and T is temperature (K)] is a straight line with

O X = 5, θ = t a n^{- 1} [- \frac{1}{2.303}]

Hence

E_{a}

and log A respectively will be:

Q. For a zero-order reaction,

A \to B

, a graph of rate vs time has a slope equal to:

[where k=rate constant]

Q. Two reactions,

A \to

Products and

B \to

Products, have rate constants

k_{A}

and

k_{B}

at temperature

T

and activation energies

E_{A}

and

E_{B}

respectively. If

k_{A} > k_{B}

and

E_{A} < E_{B}

and assuming that collision frequency for both the reactions is same, then:

The rate of reaction increase significantly with increase in temperature. Generally, rates of reactions are doubled for every

10^{\circ}

rise in temperature. Temperature coefficient gives us an idea about the change in the rate of a reaction for every

10^{\circ}

change in temperature.
Temperature coefficient

(μ) = \frac{R a t e c o n s t a n t o f (T + 10)^{\circ} C}{R a t e c o n s t a n t a t T^{\circ}}

Arrhenius gave an equation which describes rate constant k as a function of temperature

k = A e^{- E_{a} / R T}

where k is the rate constant, A is the frequency factor or pre-exponetial factor,

E_{a}

is the activation energy, T is the teperature in kelvin, and R is the universal gas constant.
Equation when expressed in logarithmic form becomes
log

k = l o g A - \frac{E_{a}}{2.303 R T}

For the given reactions, following data is given

P \to Q k_{1} = 10^{15} e x p (\frac{- 2000}{T})

C \to D k_{2} = 10^{14} e x p (\frac{- 1000}{T})

Temperature at which

k_{1} = k_{2}

is:

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The correct option is B (ii) and (v)k=Ae−Ea/kBT So if Ea=0, then k=A. also, logek=loge(A)−EakBT So, logek us 1T is a straight line.

The correct option is B (ii) and (v)
$k = A e^{- E_{a} / k_{B} T}$
So if $E_{a} = 0,$ then $k = A .$ also,
${log}_{e} k = {log}_{e} (A) - \frac{E_{a}}{k_{B} T}$
So, ${log}_{e} k$ us $\frac{1}{T}$ is a straight line.