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प्रश्न
Define the following terms:
Half-life period of reaction (t1/2).
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उत्तर
For zero-order reaction,
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संबंधित प्रश्न
In a first order reaction x → y, 40% of the given sample of compound remains unreacted in 45 minutes. Calculate rate constant of the reaction.
A → B is a first order reaction with rate 6.6 × 10-5m-s-1. When [A] is 0.6m, rate constant of the reaction is
- 1.1 × 10-5s-1
- 1.1 × 10-4s-1
- 9 × 10-5s-1
- 9 × 10-4s-1
From the rate expression for the following reaction, determine the order of reaction and the dimension of the rate constant.
\[\ce{H2O2_{( aq)} + 3I^-_{( aq)} + 2H^+ -> 2H2O_{(l)} + I^-_3}\] Rate = k[H2O2][I−]
Why molecularity is applicable only for elementary reactions and order is applicable for elementary as well as complex reactions?
Why can we not determine the order of a reaction by taking into consideration the balanced chemical equation?
Match the graph given in Column I with the order of reaction given in Column II. More than one item in Column I may link to the same item of Column II.
| Column I | Column II | |
| (i) |  |
|
| (ii) |  |
(a) 1st order |
| (iii) |  |
(b) Zero-order |
| (iv) |  |
Assertion: Order of the reaction can be zero or fractional.
Reason: We cannot determine order from balanced chemical equation.
Identify the order of reaction from the following unit for its rate constant:
L mol–1 s–1
The following data was obtained for chemical reaction given below at 975 K.
\[\ce{2NO(g) + 2H2(g) -> N2(g) + 2H2O(g)}\]
| [NO] | [H2] | Rate | |
| Mol L-1 | Mol L-1 | Mol L-1 s-1 | |
| (1) | 8 × 10-5 | 8 × 10-5 | 7 × 10-9 |
| (2) | 24 × 10-5 | 8 × 10-5 | 2.1 × 10-8 |
| (3) | 24 × 10-5 | 32 × 10-5 | 8.4 × 10-8 |
The order of the reaction with respect to NO is ______. (Integer answer)
Higher yield of NO in \[\ce{N2(g) + O2 <=> 2NO(g)}\] can be obtained at:
[ΔH of the reaction = +180.7 kJ mol−1]
- higher temperature
- lower temperature
- higher concentration of N2
- higher concentration of O2
Choose the correct answer from the options given below:
