The methane molecule, CH4, has the geometry shown in
Figure 2.17. Imagine a hypothetical process in which the
methane molecule is 'expanded,' by simultaneously extending
all four C—H bonds to infinity. We then have the
process
CH41g2¡C1g2 + 4 H1g2
(a) Compare this process with the reverse of the reaction
that represents the standard enthalpy of formation of
CH41g2.

Question

Accepted Answer

Hey everyone. So here we have the following adam Ization reaction. I'm hydrogen chloride. For the H. C. O bond is broken to produce H and cl atoms in the gas phase. They were as house this reaction similar or different from the reverse other standard entropy information of hydrogen chloride gas. So for the information we can have our elements and their standard states And it's gonna produce one mole of a substance. We're gonna have hydrogen in its standard site which is H2 gas Plus chlorine in its standard state which is cl two gas and it's gonna be one more with H. C. L. Yes, analysis balances. We have package in in the chlorine on both sides. We have two hydrogen over here to chlorine over here, one hydrogen here, one chlorine over here. Since we need to keep one mole of hydrogen chloride, We need to put one half in front of H2 and cl two. So for the reverse we're gonna have hcl yes, building one half H two yes plus one half cl two gas. So in an organization, the products are in atomic form in the reverse of the formation equation. The products are there standards form of the elements. So this is your answer. Thanks for watching my video and I hope it was helpful

Verified Solution

Key Concepts

Enthalpy of Formation

Bond Energy

Thermodynamic Processes