AB, A2 and B2 are diatomic molecules. If the bond enthalpies of A2, AB and B2 are in the ratio 2:2:1

Question

AB, A2 and B2 are diatomic molecules. If the bond enthalpies of A2, AB and B2 are in the ratio 2:2:1 and enthalpy of formation AB from A2 and B2 is −100 kJ mole–1, then what is the bond energy of A2?

Accepted Answer

$\frac{1}{2}A_{2}(g) + \frac{1}{2}B_{2}(g) \rightarrow AB(g);\Delta H = - 100\text{ kJ}$

Now, $( - 100) = \left\lbrack \frac{1}{2} \times x + \frac{1}{2} \times \frac{x}{2} \right\rbrack - x \Rightarrow x = 400$

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