The activation energy for the reaction- O3 g-NOg - NO2 g-O2 g is 9-6 KJ-mol- If - H- for this reaction is -200 KJ-mol- What is the energy of activation in KJ-mol for the reverse reaction-

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Standard X

Chemistry

Factors Affecting the Rate of a Chemical Reaction

The activatio...

Question

The activation energy for the reaction, $O_{3} (g) + N O (g) \to N O_{2} (g) + O_{2} (g) is 9.6 KJ/mol$ . If $Δ H^{\circ}$ for this reaction is $- 200 KJ/mol$ . What is the energy of activation in KJ/mol for the reverse reaction?

209.6

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209.60

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Solution

Energy of activation for reverse reaction is represented as:
$E_{r e v} = E_{f} - Δ H_{r e a c t i o n}$ ..........(1)
Given that,
$E_{f o r w a r d} = 9.6 KJ/mol$
$Δ H_{r e a c t i o n}^{\circ} = - 200 KJ/mol$
Put the values in equation (1), we get
$E_{r e v} = 9.6 - (- 200)$
$= 209.6 KJ/mol$

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The activation energy for the reaction, O3(g)+NO(g)→NO2(g)+O2(g)is 9.6 KJ/mol. If ΔH∘ for this reaction is −200 KJ/mol. What is the energy of activation in KJ/mol for the reverse reaction?

Energy of activation for reverse reaction is represented as: Erev=Ef−ΔHreaction..........(1) Given that, Eforward=9.6 KJ/mol ΔH∘reaction=−200 KJ/mol Put the values in equation (1), we get Erev=9.6−(−200) =209.6 KJ/mol

The activation energy for the reaction, $O_{3} (g) + N O (g) \to N O_{2} (g) + O_{2} (g) is 9.6 KJ/mol$ . If $Δ H^{\circ}$ for this reaction is $- 200 KJ/mol$ . What is the energy of activation in KJ/mol for the reverse reaction?

Energy of activation for reverse reaction is represented as:
$E_{r e v} = E_{f} - Δ H_{r e a c t i o n}$ ..........(1)
Given that,
$E_{f o r w a r d} = 9.6 KJ/mol$
$Δ H_{r e a c t i o n}^{\circ} = - 200 KJ/mol$
Put the values in equation (1), we get
$E_{r e v} = 9.6 - (- 200)$
$= 209.6 KJ/mol$