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Enthalpy of Combustion

For an exothe...

Question

For an exothermic reaction $A \to B$ , the activation energy is $65 k J m o l^{- 1}$ and heat of reaction $- 42 k J m o l^{- 1}$ . The activation energy for the reaction $B \to A$ would be :-

23 k J m o l^{- 1}

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107 k J m o l^{- 1}

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65 k J m o l^{- 1}

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42 k J m o l^{- 1}

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Solution

The correct option is B $107 k J m o l^{- 1}$
$A \to B$
$(E_{a})_{1} = 65 k J / m o l$
$△ H = - 42 k J / m o l$
Now, For $B \to A$
$(E_{a})_{2} = (E_{a})_{1} - △ H$
$= 65 - (- 42) = 107 k J / m o l$

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For an exothermic reaction A→B, the activation energy is 65 kJ mol−1 and heat of reaction −42 kJ mol−1. The activation energy for the reaction B→A would be :-

The correct option is B 107 kJ mol−1A→B(Ea)1=65 kJ/mol△H=−42 kJ/molNow, For B→A(Ea)2=(Ea)1−△H=65−(−42)=107 kJ/mol

For an exothermic reaction $A \to B$ , the activation energy is $65 k J m o l^{- 1}$ and heat of reaction $- 42 k J m o l^{- 1}$ . The activation energy for the reaction $B \to A$ would be :-

The correct option is B $107 k J m o l^{- 1}$
$A \to B$
$(E_{a})_{1} = 65 k J / m o l$
$△ H = - 42 k J / m o l$
Now, For $B \to A$
$(E_{a})_{2} = (E_{a})_{1} - △ H$
$= 65 - (- 42) = 107 k J / m o l$