A first order reaction completed its 20% in 200 minute. How much

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 Multiple Choice QuestionsMultiple Choice Questions

471.

Half-life period of a radioactive element is 100 yr. How long will it take for its 93. 75% decay?

  • 400yr

  • 300yr

  • 200yr

  • 193yr


472.

The unit of rate constant of a second order reaction is

  • mol/L-s

  • L/mol-s

  • L2/mol2-s

  • per second


473.

2N2O5  4NO2 + O2

For the above reaction which of the following is not correct about rates of reaction?

  • -d[N2O5]dt= 2d[O2]dt

  • -2d[N2O5]dt= d[NO2]dt

  • -d[NO2]dt= 4d[O2]dt

  • -2d[N2O5]dt= 4d[NO2]dt= d[O2]dt


474.

A + B → Product

If concentration of A is doubled, rate increases 4 times. If concentrations of A and B both are doubled, rate increases 8 times. The differential rate equation of the reaction will be

  • dCdt= kCA× CB

  • dCdt= kCA2 × CB3

  • dCdt= kCA2 × CB

  • dCdt= kCA2 × CB2


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475.

Which of the following is a first order reaction?

  • NH4NO2 →N2 + 2H2O

  • 2HI → H2 + I2

  • 2NO2 → 2NO + O2

  • 2NO + O2 → 2NO2


476.

In the reaction of A + 2B → C + 2D, the initial rate = -d[A]dt att = 0 wasfound to be 2.6 × 10-2 M s-1. What is the value of -d[B]dt at t = 0 in Ms-1 ?

  • 2.6 × 10-2

  • 5.2 × 10-2

  • 1.0 × 10-1

  • 6.5 × 10-3


477.

The thermal decomposition of a compound is of first order. If a sample of the compound decomposes 50% in 120 min, what time will it take to undergo 90% decomposition?

  • Nearly 400 min

  • Nearly 45 min

  • Nearly 480 min

  • Nearly 240 min


478.

What is the half-life of Na-24, if 2 x10-4 g sample of it disintegrates at the rate of 7.0 x 1012  atoms per second?

  • 4.97 × 106

  • 4.97 × 105

  • 0.497 × 105 s

  • 4.97 × 104 s


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479.

Energy of activation of an endothermic reaction is

  • negative

  • positive

  • zero

  • Cannot be predicted


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480.

A first order reaction completed its 20% in 200 minute. How much time it will take to complete its 80%?

  • 400 min

  • 800 min

  • 1400 min

  • 1000 min


C.

1400 min

For the first order reaction,

k = 2.303t log aa - x

For 20% completion of the reaction,

k = 2.303200 log aa - x   = 2.303200 log 1.25   = 2.303 × 0.0969200

For 80% completion of the reaction,

k = k = 2.303t log 100100 - 80    = 2.303t log   = 2.303 × 0.699t

On putting the value of k, we get

2.303 × 0.0969200 = 2.303 × 0.699t t = 200 × 0.6990.0969

      = 1442 min ≈ 1400 min


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