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प्रश्न
The reaction between \[\ce{H2(g)}\] and \[\ce{O2(g)}\] is highly feasible yet allowing the gases to stand at room temperature in the same vessel does not lead to the formation of water. Explain.
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उत्तर
\[\ce{2H2(g) + O2(g) -> 2H2O(I)}\]
This reaction does not take place at room because the activation energy of the reaction is very high.
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संबंधित प्रश्न
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`3I_((aq))^-) +S_2O_8^(2-)->I_(3(aq))^-) + 2S_2O_4^(2-)`
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c. `-(d[I_3^-])/dt`
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(i) energy is always released
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| Column I | Column I | |
| (i) | Catalyst alters the rate of reaction | (a) cannot be fraction or zero |
| (ii) | Molecularity | (b) proper orientation is not there always |
| (iii) | Second half life of first order reaction | (c) by lowering the activation energy |
| (iv) | `e^((-E_a)/(RT)` | (d) is same as the first |
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| (vi) | Area under the Maxwell Boltzman curve is constant | (f) refers to the fraction of molecules with energy equal to or greater than activation energy |
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