For the following reaction scheme (homogeneous), the rate constant has units : \(A + B \overset{k}{\rightarrow} C\)

sec^-1 mole
sec^-1
sec^-1 litre mole^-1
sec

The first order rate constant for a certain reaction increases from 1.667 × 10^-6 s^-1 at 727ºC to 1.667 × 10^-4 s^-1 at 1571 ºC.The rate constant at 1150ºC, assuming constancy of activation energy over the given temperature range is[Given : log 19.9 = 1.299]

3.911 × 10^-5 s^-1
1.139 × 10^-5 s^-1
3.318 × 10^-5 s^-1
1.193 × 10^-5 s^-1

Solution

Velocity constant K of are action is affected by

Change in the concentration of the reactant
Change of temperature
Change in the concentration of the product
None of the above

Solution

Velocity constant ‘K’ is characteristic constant of are action and depends only on temperature and catalyst.

T₅₀ of first -order reaction is 10 min. Starting with 10 mol L^-1,rate after 20 minis

0.0693 mol L^-1 min^-1
0.0693 × 2.5 mol L^-1min^-1
0.0693 × 5 mol L^-2 min^-1
0.0693 × 10 mol L^-1 min^-1

Solution

The rate constant for the reaction 2N₂O₅ ⟶ 4NO₂ + O₂ is 3.10 ×10^-5 sec–1. If the rate is 2.4 ×10^-5 mol litre^-1 sec^-1 then the concentration of N₂O₅ (in mol litre^-1) is :

0.04
0.8
0.07
1.4

Solution

In a first order reaction the concentration of reactant decreases from 800 mol/dm³ to 50 mol/dm³ in 2 × 104 sec.The rate constant of reaction in sec^-1 is:

2 × 10⁴
3.45 × 10^-5
1.386 × 10^-4
2 × 10^-4

Solution

For a first order reaction, to obtain a positive slope, we need to plot {where [A] is the concentration of reactant A}

– log₁₀[A] vs t
– log_e[A] vs t
log₁₀[A] vs log t
[A] vs t

Solution

For a first order reaction the positive slope is obtained when we plot – log_e[A] vs t.

For the reaction 2N₂O₅ ⟶ 4NO₂ + O₂,rate and rate constant are 1.02× 10^-4 mol lit^-1sec^-1 and 3.4 × 10^-5 sec^-1 respectively then concentration of N₂O₅ at that time will be

1.732M
3M
3.4 × 10⁵ M
1.02 × 10^-4 M

Solution

Solution

From data 1 and 3, it is clear that keeping (B) const, [A] is doubled, rate remains unaffected. Hence rate is independent of [A]. From 1 and 4, keeping [A] constant,[B] is doubled, rate become 8 times. Hence rate ∝ [B]³

Solution

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