What fraction of a reactant in first order reaction is left after 40 minute if 1 2 t is 20 minute

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  • Physical Chemistry
  • chemical kinetics

DEEPANKAR TULSIANI,

12 years ago

Grade:12

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1 Answers

aakanksha sharma

4 Points

12 years ago

hi deepankar,

ans cannot be d in any case.listen if t1/2 is 20 min then after 40 min. 1/4th of initial amt.will be left.then ans. is a.

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  1. NEET Questions


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Subtopic:  First Order Reaction Kinetics |

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NEET 2023 - Target Batch - Aryan Raj Singh

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Rate equation for a second order reaction is:

1. K = (2.303/t) log {a/(a-x)}

2. K = (1/t) log {a/(a-x)}

3. K = (1(a)t-1(ao)t)* 1t

4. K = (1/t2) log {a/(a-x)}

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Subtopic:  Order,Molecularity and Mechanism |

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For the reaction 2NO2 + F2 → 2NO2F, following

mechanism has been provided,

 NO2 + F2   →slow  NO2F+F

NO2 + F   →fast NO2F

Thus, rate expression of the above

reaction can be written as:

1. r = K[NO2]2[F2]

2. r = K[NO2 ][F2]

3. r = K[NO2]

4. r = K[F2]

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Subtopic:  Definition, Rate Constant, Rate Law |

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For the reaction:

[Cu(NH3)4]2+ + H2O⇌[Cu(NH3)3H2O]2+ + NH3

the net rate of reaction at any time is given by, net rate =

2.0x10-4 [Cu(NH3)4]2+[H2O] - 3.0x105 [Cu(NH3 )3 H20]2+[NH3]

Then correct statement is/are :

1. rate constant for forward reaction = 2 x 10-4

2. rate constant for backward reaction = 3 x 105

3. equilibrium constant for the reaction = 6.6 x 10-10

4. all of the above

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Subtopic:  Definition, Rate Constant, Rate Law |

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Rate constant of reaction can be expressed by Arrhenius equation as,

                             K= Ae-EaRT   

In this equation, Ea represents:

1. the energy above which all the colliding molecules will react

2. the energy below which colliding molecules will not react

3. the total energy of the reacting molecules at a temperature, T

4. the fraction of molecules with energy greater than the activation energy of the reaction

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Subtopic:  Arrhenius Equation |

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Consider the chemical reaction,

N2g+3H2g→2NH3g

The rate of this reaction can be expressed in terms of time derivative of concentration of  N2 g, H2g and NH3g.

The correct relationship amongest the rate expressions is: 

(1) Rate =-dN2dt=-13 dH2 dt=12 dNH3dt

(2) Rate =-dN2 dt=-3 dH2dt=2 dNH3dt

(3) Rate =dN2dt=13 dH2dt =12 dNH3dt

(4) Rate =-dN2dt =- dH2dt= dNH3dt

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Subtopic:  Definition, Rate Constant, Rate Law |

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For a first order reaction A→ Product, the initial concentration of A is 0.1 M and after 40 minute it becomes 0.025 M. Calculate the rate of reaction at reactant concentration of 0.01M:

1. 3.47x10-4 M min-1

2. 3.47x10-5 M min-1

3. 1.735 x 10-6 M min-1

4. 1.735 x10-4 M min-1

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Subtopic:  First Order Reaction Kinetics |

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Select the intermediate in the following reaction mechanism:

O3(g) ⇌ O2(g) +O(g)

O(g) +O3(g) → 2O2(g)

1. O3(g)

2. O(g)

3. O2(g)

4. none of these

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Subtopic:  Definition, Rate Constant, Rate Law |

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A reactant with initial concentration 1.386 mol litre-1 showing first order change takes 40 minute to become half. If it shows zero order change taking 20 minute to becomes half under the similar conditions, the ratio, K1/K0 for first order and zero order kinetics will be:

1. 0.5 mol-1 litre

2. 1.0 mol/litre

3. 1.5 mol/litre

4. 2.0 mol-1 litre

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Subtopic:  First Order Reaction Kinetics |

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In a first order reaction, the concentration of the reactant is decreased from 1.0 M to 0.25M in 20 minute. The rate constant of the reaction would be:

1. 10min-1

2. 6.931 min-1

3. 0.6931 min-1

4. 0.06931 min-1

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  • (2)

  • (3)

  • (4)

Subtopic:  First Order Reaction Kinetics |

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What fraction of a reactant in first order reaction is left after 40 minute?

0.69340=2.303160log[1(a−x)];(a−x)=1/16.

What fraction of a reactant in first order reaction is left after 40 minutes if half is 20 minutes?

a−xa=41.

How do you calculate time for a first order reaction?

Calculation for the time taken for a first order reaction to complete to 75% (using log table)- Chemical Kinetics.
k = 2.303/20 . log (a/¾ a) = 2.303/20. ... .
Similarly, for 75% (¾ of a) of the starting material to decompose, remaining starting material will be ¼ of a..
t= 2.303/k . log (a/a-x) = 2.303/k ..

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