Answer:
∆H° rxn = - 93 kJ
Explanation:
Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds broken minus bonds formed (H according to Hess Law.
We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.
N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)
1 N≡N = 1(945 kJ/mol) 3 H-H = 3 (432 kJ/mol) 6 N-H = 6 ( 389 kJ/mol)
∆H° rxn = ∑ H bonds broken - ∑ H bonds formed
∆H° rxn = [ 1(945 kJ) + 3 (432 kJ) ] - [ 6 (389 k J]
∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ
be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond, N≡N, we have to use this one .
The quantity of heat required to vapourize 1 mole of a substance depends on the kind of intermolecular forces between the molecules of the substance. Diethyl ether molecules are held together by weak dispersion forces compared to the stronger hydrogen bonding in ethanol. Therefore, 1 mole of diethyl ether requires less heat to vapourize than is required to vapourize 1 mole of ethanol.
Intermolecular forces hold the molecules a substance together in a given state of matter. The properties of a substance such as boiling point, melting point etc are dependent on the nature of intermolecular forces holding the molecules of the substance.
Diethyl ether molecules are held together by weak dispersion forces while molecules of ethanol are held together by hydrogen bonds.
Since hydrogen bonds are much stronger than dispersion forces, a greater quantity of heat is required to break the intermolecular hydrogen bonds in ethanol in order to vapourize them than is required to vapourize diethyl ether.
Therefore, owing to stronger intermolecular forces between molecules of ethanol, less heat is required to vapourize than is required to vapourize 1 mole of ethanol.
Learn more: brainly.com/question/9328418
I believe that the answer is B
I think it’s A sorry if i’m wrong
