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Standard VII

Chemistry

Activation Energy

For a reactio...

Question

For a reaction $A \to B$ , enthalpy of reaction is $- 4.2 k J m o l^{- 1}$ and enthalpy of activation is $9.6 k J m o l^{- 1}$ . The correct potential energy profile for the reaction is shown in option.

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Solution

The correct option is C

Enthaply of Reaction, $Δ H_{r x n} = (E_{a})_{f} - (E_{a})_{b}$

$- 4.2 = (E_{a})_{f} - (E_{a})_{b}$

$- 4.2 = 9.6 - (E_{a})_{b}$
Activation Energy Of Backword Reaction, $(E_{a})_{b} = 9.6 + 4.2 = 13.8 k J m o l^{- 1}$

Since reaction is exothermic, so possible graph is (C) only.

Also $(E_{a})_{f} < (E_{a})_{b},$ so answer is option (C).

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For a reaction A→B, enthalpy of reaction is –4.2 kJ mol–1 and enthalpy of activation is 9.6 kJ mol–1. The correct potential energy profile for the reaction is shown in option.

The correct option is C Enthaply of Reaction, ΔHrxn=(Ea)f−(Ea)b −4.2=(Ea)f−(Ea)b −4.2=9.6−(Ea)b Activation Energy Of Backword Reaction, (Ea)b=9.6+4.2=13.8 kJ mol−1 Since reaction is exothermic, so possible graph is (C) only. Also (Ea)f<(Ea)b, so answer is option (C).

For a reaction $A \to B$ , enthalpy of reaction is $- 4.2 k J m o l^{- 1}$ and enthalpy of activation is $9.6 k J m o l^{- 1}$ . The correct potential energy profile for the reaction is shown in option.