1e:I[80107,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 1f:I[76763,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 1d:T441,{"@context":"https://schema.org","@graph":[{"@type":"BreadcrumbList","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https://www.divinejee.com"},{"@type":"ListItem","position":4,"name":"A2B2 arrow 2 AB Delta Hf0200 kJ mol1 AB A2 and B2 are diatom","item":"https://www.divinejee.com/question/a2-b2-arrow-2-ab-delta-hf0200-kj-mol1-ab-a2-and-b2-are-diatomic-molecules-if-the-bond-enthalpies"}]},{"@type":"QAPage","mainEntity":{"@type":"Question","name":"Question A2B2 arrow 2 AB Delta Hf0200 kJ mol1 AB A2 and B2 are diatom on Topic","text":"

$\\mathrm{A}_{2}+\\mathrm{B}_{2} \\rightarrow 2 \\mathrm{AB} . \\Delta H_{f}^{0}=-200 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$

$\\mathrm{AB}, \\mathrm{A}_{...","answerCount":1,"upvoteCount":0,"datePublished":"2026-01-16","author":{"@type":"Organization","name":"DivineJEE"},"acceptedAnswer":{"@type":"Answer","text":"Check the detailed solution on the page.","upvoteCount":0,"url":"https://www.divinejee.com/question/a2-b2-arrow-2-ab-delta-hf0200-kj-mol1-ab-a2-and-b2-are-diatomic-molecules-if-the-bond-enthalpies"}}}]}20:T4eb,To find the bond enthalpy of $$\mathrm{A}_{2}$$, we can use the given information about the reaction and the bond enthalpies' ratio. The reaction is:

$$\mathrm{A}_{2}+\mathrm{B}_{2} \rightarrow 2 \mathrm{AB}$$

The enthalpy change for the reaction, $$\Delta H_{f}^{0}$$, is given as:

$$\Delta H_{f}^{0} = -200 \mathrm{~kJ} \mathrm{~mol}^{-1}$$

The bond enthalpies of $$\mathrm{A}_{2}$$, $$\mathrm{B}_{2}$$, and $$\mathrm{AB}$$ are in the ratio of $$1: 0.5: 1$$.

Let's denote the bond enthalpies of $$\mathrm{A}_{2}$$, $$\mathrm{B}_{2}$$, and $$\mathrm{AB}$$ as $$x$$, $$0.5x$$, and $$x$$, respectively.

The enthalpy change for the reaction can be calculated using the bond enthalpies:

$$\Delta H_{f}^{0} = \text{(Sum of bond enthalpies of reactants)} - \text{(Sum of bond enthalpies of products)}$$

For the given reaction:

$$-200 = (x + 0.5x) - 2x$$

Now we can solve for $$x$$, which represents the bond enthalpy of $$\mathrm{A}_{2}$$:

$$-200 = 1.5x - 2x$$

$$-200 = -0.5x$$

$$x = \frac{-200}{-0.5}$$

$$x = 400$$

The bond enthalpy of $$\mathrm{A}_{2}$$ is $$400 ~\mathrm{kJ} ~\mathrm{mol}^{-1}$$.