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प्रश्न
What is pseudo first order reaction? Give one· example of it.
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उत्तर
A reaction which has higher order true rate law but experimentally found to behave as first order are called pseudo first order reaction.
e.g. `C_12H_22O_(11(aq))+H_2OrarrC_6H_12O_6+C_6H_12O_6`
Sucrose Glucose fractose
is an example of pseudo first order reaction, because water takes part in the reaction the true rate law
rate = k'[C12H22O11][H2O]
indicates that the reaction must be second order. Similarly to the hydrolysis of ester,[H2O] is constant and
the rate law became
rate=k[C12H22O11]
Thus, the second order true law is converted, to first order rate law.
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संबंधित प्रश्न
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| t/s | 0 | 30 | 60 |
| [CH3COOCH3] / mol L–1 | 0.60 | 0.30 | 0.15 |
(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
(ii) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(Given log 2 = 0.3010, log 4 = 0.6021)
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SO2Cl2 (g) → SO2 (g) + Cl2 (g)
| Experiment | Time/s–1 | Total pressure/atm |
| 1 | 0 | 0.4 |
| 2 | 100 | 0.7 |
Calculate the rate constant.
(Given : log 4 = 0.6021, log 2 = 0.3010)
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\[\ce{2A + B → A2B}\]
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| A/mol L−1 | 0.20 | 0.20 | 0.40 |
| B/mol L−1 | 0.30 | 0.10 | 0.05 |
| r0/mol L−1 s−1 | 5.07 × 10−5 | 5.07 × 10−5 | 1.43 × 10−4 |
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Reason (R): For a complex reaction, molecularity has no meaning.
A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is reduced to half?
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:
