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CHXII04:CHEMICAL KINETICS

320274 The rate constant $k_{1}$ of a reaction is found to be double that of rate constant $k_{2}$ of another reaction. The relationship between corresponding activation energies of the two reactions at same temperature $(E_{1} a n d E_{2})$ can be represented as

1

E_{1} > E_{2}

2

E_{1} < E_{2}

3

E_{1} = E_{2}

4 none

Explanation:

$k_{1} = A_{1} e^{- E_{1} / R T} k_{2} = A_{2} e^{- E_{2} / R T}$
$\frac{k_{1}}{k_{2}} = \frac{A_{1}}{A_{2}} \times e^{- (E_{1} + E_{2}) / R T}; A_{1} a n d A_{2} a r e n o t$ given

CHXII04:CHEMICAL KINETICS

320275 The activation energy of a reaction at a given temperature is found to be $2 .303 RT J mo l^{- 1}$ . The ratio of rate constant $(k)$ to the arrhenius factor (A) is:

1 0.01

2 0.1

3 0.02

4 0.001

Explanation:

$k = {Ae}^{- Ea / RT}$
$\log k = \log A - \frac{Ea}{RT \times 2.303}$
$\log k - \log A = \frac{- Ea}{2.303 RT}$
$\log \frac{k}{A} = \frac{- Ea}{2.303 RT} = \frac{- 2.303 RT}{2.303 RT}$
$\frac{k}{A} = anti \log - 1 = 1 \times 10^{- 1} = 0.1$

CHXII04:CHEMICAL KINETICS

320276 Activation energy of a reaction is:

1 the energy released during the reaction

2 the energy evolved when activated complex is formed

3 additional amount of energy needed to overcome the potential barrier of reaction

4 the energy needed to form one mole of the product

CHXII04:CHEMICAL KINETICS

320277 The rate constant for the first order decomposition of the ethylene oxide into ${CH}_{4}$ and $CO$ is described by $\log k (s^{- 1}) = 14.34 - \frac{12.5 \times 10^{4} K}{T}$
Then the activation energy of the reaction is
supporting img

1

2.39 \times 10^{5} kJ {mol}^{- 1}

2

2.39 \times 10^{3} kJ {mol}^{- 1}

3

4.78 \times 10^{5} kJ {mol}^{- 1}

4

4.78 \times 10^{2} kJ {mol}^{- 1}

Explanation:

$\log k = \log A - \frac{E_{a}}{2.303 RT}$
$\log k = 14.34 - \frac{12.5 \times 10^{4} K}{T}$
From Eqs. (1) and (2)
$\frac{E_{a}}{2.303 R} = 12.5 \times 10^{4} K$
$E_{a} = (12.5 \times 10^{4} K) \times 2.303 \times 8.314 {JK}^{- 1} {mol}^{- 1}$
$= 239 \times 10^{4} J {mol}^{- 1}$
$= 2.39 \times 10^{6} J {mol}^{- 1}$
$= 2.39 \times 10^{3} kJ {mol}^{- 1}$

CHXII04:CHEMICAL KINETICS

320274 The rate constant $k_{1}$ of a reaction is found to be double that of rate constant $k_{2}$ of another reaction. The relationship between corresponding activation energies of the two reactions at same temperature $(E_{1} a n d E_{2})$ can be represented as

1

E_{1} > E_{2}

2

E_{1} < E_{2}

3

E_{1} = E_{2}

4 none

Explanation:

$k_{1} = A_{1} e^{- E_{1} / R T} k_{2} = A_{2} e^{- E_{2} / R T}$
$\frac{k_{1}}{k_{2}} = \frac{A_{1}}{A_{2}} \times e^{- (E_{1} + E_{2}) / R T}; A_{1} a n d A_{2} a r e n o t$ given

CHXII04:CHEMICAL KINETICS

320275 The activation energy of a reaction at a given temperature is found to be $2 .303 RT J mo l^{- 1}$ . The ratio of rate constant $(k)$ to the arrhenius factor (A) is:

1 0.01

2 0.1

3 0.02

4 0.001

Explanation:

$k = {Ae}^{- Ea / RT}$
$\log k = \log A - \frac{Ea}{RT \times 2.303}$
$\log k - \log A = \frac{- Ea}{2.303 RT}$
$\log \frac{k}{A} = \frac{- Ea}{2.303 RT} = \frac{- 2.303 RT}{2.303 RT}$
$\frac{k}{A} = anti \log - 1 = 1 \times 10^{- 1} = 0.1$

CHXII04:CHEMICAL KINETICS

320276 Activation energy of a reaction is:

1 the energy released during the reaction

2 the energy evolved when activated complex is formed

3 additional amount of energy needed to overcome the potential barrier of reaction

4 the energy needed to form one mole of the product

CHXII04:CHEMICAL KINETICS

320277 The rate constant for the first order decomposition of the ethylene oxide into ${CH}_{4}$ and $CO$ is described by $\log k (s^{- 1}) = 14.34 - \frac{12.5 \times 10^{4} K}{T}$
Then the activation energy of the reaction is
supporting img

1

2.39 \times 10^{5} kJ {mol}^{- 1}

2

2.39 \times 10^{3} kJ {mol}^{- 1}

3

4.78 \times 10^{5} kJ {mol}^{- 1}

4

4.78 \times 10^{2} kJ {mol}^{- 1}

Explanation:

$\log k = \log A - \frac{E_{a}}{2.303 RT}$
$\log k = 14.34 - \frac{12.5 \times 10^{4} K}{T}$
From Eqs. (1) and (2)
$\frac{E_{a}}{2.303 R} = 12.5 \times 10^{4} K$
$E_{a} = (12.5 \times 10^{4} K) \times 2.303 \times 8.314 {JK}^{- 1} {mol}^{- 1}$
$= 239 \times 10^{4} J {mol}^{- 1}$
$= 2.39 \times 10^{6} J {mol}^{- 1}$
$= 2.39 \times 10^{3} kJ {mol}^{- 1}$

CHXII04:CHEMICAL KINETICS

320274 The rate constant $k_{1}$ of a reaction is found to be double that of rate constant $k_{2}$ of another reaction. The relationship between corresponding activation energies of the two reactions at same temperature $(E_{1} a n d E_{2})$ can be represented as

1

E_{1} > E_{2}

2

E_{1} < E_{2}

3

E_{1} = E_{2}

4 none

Explanation:

$k_{1} = A_{1} e^{- E_{1} / R T} k_{2} = A_{2} e^{- E_{2} / R T}$
$\frac{k_{1}}{k_{2}} = \frac{A_{1}}{A_{2}} \times e^{- (E_{1} + E_{2}) / R T}; A_{1} a n d A_{2} a r e n o t$ given

CHXII04:CHEMICAL KINETICS

320275 The activation energy of a reaction at a given temperature is found to be $2 .303 RT J mo l^{- 1}$ . The ratio of rate constant $(k)$ to the arrhenius factor (A) is:

1 0.01

2 0.1

3 0.02

4 0.001

Explanation:

$k = {Ae}^{- Ea / RT}$
$\log k = \log A - \frac{Ea}{RT \times 2.303}$
$\log k - \log A = \frac{- Ea}{2.303 RT}$
$\log \frac{k}{A} = \frac{- Ea}{2.303 RT} = \frac{- 2.303 RT}{2.303 RT}$
$\frac{k}{A} = anti \log - 1 = 1 \times 10^{- 1} = 0.1$

CHXII04:CHEMICAL KINETICS

320276 Activation energy of a reaction is:

1 the energy released during the reaction

2 the energy evolved when activated complex is formed

3 additional amount of energy needed to overcome the potential barrier of reaction

4 the energy needed to form one mole of the product

CHXII04:CHEMICAL KINETICS

320277 The rate constant for the first order decomposition of the ethylene oxide into ${CH}_{4}$ and $CO$ is described by $\log k (s^{- 1}) = 14.34 - \frac{12.5 \times 10^{4} K}{T}$
Then the activation energy of the reaction is
supporting img

1

2.39 \times 10^{5} kJ {mol}^{- 1}

2

2.39 \times 10^{3} kJ {mol}^{- 1}

3

4.78 \times 10^{5} kJ {mol}^{- 1}

4

4.78 \times 10^{2} kJ {mol}^{- 1}

Explanation:

$\log k = \log A - \frac{E_{a}}{2.303 RT}$
$\log k = 14.34 - \frac{12.5 \times 10^{4} K}{T}$
From Eqs. (1) and (2)
$\frac{E_{a}}{2.303 R} = 12.5 \times 10^{4} K$
$E_{a} = (12.5 \times 10^{4} K) \times 2.303 \times 8.314 {JK}^{- 1} {mol}^{- 1}$
$= 239 \times 10^{4} J {mol}^{- 1}$
$= 2.39 \times 10^{6} J {mol}^{- 1}$
$= 2.39 \times 10^{3} kJ {mol}^{- 1}$

CHXII04:CHEMICAL KINETICS

320274 The rate constant $k_{1}$ of a reaction is found to be double that of rate constant $k_{2}$ of another reaction. The relationship between corresponding activation energies of the two reactions at same temperature $(E_{1} a n d E_{2})$ can be represented as

1

E_{1} > E_{2}

2

E_{1} < E_{2}

3

E_{1} = E_{2}

4 none

Explanation:

$k_{1} = A_{1} e^{- E_{1} / R T} k_{2} = A_{2} e^{- E_{2} / R T}$
$\frac{k_{1}}{k_{2}} = \frac{A_{1}}{A_{2}} \times e^{- (E_{1} + E_{2}) / R T}; A_{1} a n d A_{2} a r e n o t$ given

CHXII04:CHEMICAL KINETICS

320275 The activation energy of a reaction at a given temperature is found to be $2 .303 RT J mo l^{- 1}$ . The ratio of rate constant $(k)$ to the arrhenius factor (A) is:

1 0.01

2 0.1

3 0.02

4 0.001

Explanation:

$k = {Ae}^{- Ea / RT}$
$\log k = \log A - \frac{Ea}{RT \times 2.303}$
$\log k - \log A = \frac{- Ea}{2.303 RT}$
$\log \frac{k}{A} = \frac{- Ea}{2.303 RT} = \frac{- 2.303 RT}{2.303 RT}$
$\frac{k}{A} = anti \log - 1 = 1 \times 10^{- 1} = 0.1$

CHXII04:CHEMICAL KINETICS

320276 Activation energy of a reaction is:

1 the energy released during the reaction

2 the energy evolved when activated complex is formed

3 additional amount of energy needed to overcome the potential barrier of reaction

4 the energy needed to form one mole of the product

CHXII04:CHEMICAL KINETICS

320277 The rate constant for the first order decomposition of the ethylene oxide into ${CH}_{4}$ and $CO$ is described by $\log k (s^{- 1}) = 14.34 - \frac{12.5 \times 10^{4} K}{T}$
Then the activation energy of the reaction is
supporting img

1

2.39 \times 10^{5} kJ {mol}^{- 1}

2

2.39 \times 10^{3} kJ {mol}^{- 1}

3

4.78 \times 10^{5} kJ {mol}^{- 1}

4

4.78 \times 10^{2} kJ {mol}^{- 1}

Explanation:

$\log k = \log A - \frac{E_{a}}{2.303 RT}$
$\log k = 14.34 - \frac{12.5 \times 10^{4} K}{T}$
From Eqs. (1) and (2)
$\frac{E_{a}}{2.303 R} = 12.5 \times 10^{4} K$
$E_{a} = (12.5 \times 10^{4} K) \times 2.303 \times 8.314 {JK}^{- 1} {mol}^{- 1}$
$= 239 \times 10^{4} J {mol}^{- 1}$
$= 2.39 \times 10^{6} J {mol}^{- 1}$
$= 2.39 \times 10^{3} kJ {mol}^{- 1}$

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