What is the activation energy for reverse reaction on the basis of given data ?

N₂O₄(g) $\to$ 2NO₂(g)  ΔH = +54kJ

E_a(forward) = +57.2 kJ

1. -54 kJ

2.  +3.2 kJ

3. 60.2 kJ

4. 111.2 kJ

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

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

1. increases with an increase in temperature

2. decreases with a decrease in temperature

3. decreases with an increase in temperature

4. is nearly independent of temperature

If 2M, 1L solution of acetic acid is added to 3M, 1L ethyl alcohol then the following elementary reaction takes place.

CH₃COOH + C₂H₅OH $\to$ CH₃COOC₂H₅ + H₂0

If each solution is diluted by 1 litre then the initial rate becomes how many times:-

(1) 4

(2) 2

(3) 0.5

(4) 0.25

When temperature is increased from 27$°$C to 127$°$C, rate of reaction becomes doubled, then E_a will be?

1. 1.66 kcal.

2. 3.32 kcal.

3. 5.33 kcal.

4. 6.64 kcal.

The dissociation of H₂O₂ is a first order reaction. The half life for '16 V H₂O₂ is 30 min, calculate the time at which the solution is 1V H₂O₂ ?

(1) 120 min.

(2) 90 min.

(3) 60 min.

(4) 150 min.

The correct expression for the 3/4th life of a first-order reaction is:
\(1 .  \frac{k}{2 . 303} log \frac{4}{3}\)
\(2 .  \frac{2 . 303}{k} log \frac{3}{4}\)
\(3 .  \frac{2 . 303}{k} log\) \(4\)
\(4 .  \frac{2 . 303}{k}\) \(log\) \(3\)$\mathrm{}$

The activation energies of the forward and backward reactions in the case of a chemical reaction are 30.5 and 45.4 KJ/mol respectively. The reaction is

1. Exothermic

2. Endothermic

3. Neither exothermic nor endothermic

4. Independent of temperature

Mechanism of a hypothetical reaction

${\mathrm{X}}_2<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{Y}}_2<mspace\; width="1em"></mspace>\stackrel{<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>}{\to }<mspace\; width="1em"></mspace>2\mathrm{XY}$ is given below

(i) ${\mathrm{X}}_2<mspace\; width="1em"></mspace>\stackrel{<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>}{\rightleftharpoons }<mspace\; width="1em"></mspace>\mathrm{X}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\mathrm{X}<mspace\; width="1em"></mspace>\left(\mathrm{fast}\right)$

(ii) $\mathrm{X}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{Y}}_2<mspace\; width="1em"></mspace>\stackrel{<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>}{\to }<mspace\; width="1em"></mspace>\mathrm{XY}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\mathrm{Y}<mspace\; width="1em"></mspace>\left(\mathrm{slowest}\right)$

(iii) $\mathrm{X}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\mathrm{Y}<mspace\; width="1em"></mspace>\stackrel{<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>}{\to }\mathrm{XY}<mspace\; width="1em"></mspace>\left(\mathrm{fast}\right)$

The overall order of the reaction will be

1. 1

2.  2

3. 0

4. 1.5

The activation energy of a reaction can be determined from the slope of the graph between : 

1. ln K vs T

2. ln $\frac{\mathrm{K}}{\mathrm{T}}$ vs T

3. ln K vs $\frac{1}{\mathrm{T}}$

4. $\frac{\mathrm{T}}{\ln <mspace\; width="1em"></mspace>\mathrm{K}}<mspace\; width="1em"></mspace>vs<mspace\; width="1em"></mspace>\frac{1}{T}$