The reaction between $H_{2}$ and $O_{2}$ is highly feasible yet allowing the gases to stand at room temperature in the same vessel does not lead to the formation of $H_{2} O$ .

True

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False

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Solution

The correct option is A True
Solution:
This is because Because activation energy of the reaction is very high at room temperature
but at high-temperature,
$H_{2} (g)$ and $O_{2} (g)$
bond break and colliding particles cross the energy barrier. This is why the reaction between $H_{2} (g)$ and $O_{2} (g)$ does not lead to the formation of water at room temperature while keeping in the same vessel.
hence this statement is true
hence the correct opt: A

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Similar questions

The reaction between $H_{2} (g)$ and $O_{2} (g)$ 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.

Q. The reaction

C r_{2} O_{3} + 2 A l \to A l_{2} O_{3} + 2 C r (△ G^{0} = - 421 k J)

is thermodynamically feasible due to -ve value of

△ G .

Why does this reaction not take place at room temperature?

Q. If hydrogen and oxygen are mixed and kept in the same vessel at room temperature, the reaction does not take place to form water because:

Q. At room temperature, the reaction between

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and

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to give

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is fast, while that between

C O

and

O_{2}

is slow. It is due to:

Q. A vessel contains

H_{2}

and

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in the molar ratio of 8 : 1. This mixture of gases is allowed to diffuse through a hole. Find the ratio of the composition of

H_{2}

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coming out through the hole.

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The reaction between H2 and O2 is highly feasible yet allowing the gases to stand at room temperature in the same vessel does not lead to the formation of H2O.

The reaction between $H_{2}$ and $O_{2}$ is highly feasible yet allowing the gases to stand at room temperature in the same vessel does not lead to the formation of $H_{2} O$ .