A. N₂ (g) + 3 H₂ (g) --> 2 NH₃ (g)
B. The value for standard enthalpy of formation is empirical given that the reactants involved were pure elements. So, you can search this on the internet or in any textbook. The Hf for NH₃ is -46.0 kJ/mol.
C. C (s) + O₂ (g) --> CO₂ (g)
D. The Hf for CO₂ is <span>-393.5 kJ/mol
E. 4 Fe (s) + 3 O</span>₂ (g) --> 2 Fe₂O₃ (s)
F. The Hf for solid Fe₂O₃ is -826.0 kJ/mol.
G. C (s) + 2 H₂ (g) --> CH₄ (g)
H. The Hf for methane gas is -74.9 kJ/mol.
14 as well. A neutral atom has just as many protons in its nucleus and electrons it is electronic shells.
Answer:
Explanation:
Concepts and reason
This problem is based on the concept of hydrolysis of esters.
An ester is hydrolyzed to a carboxylic acid and an alcohol when treated with aqueous acid or aqueous base. Under alkaline conditions, the carboxylic acid is obtained in the form as its salt.
Fundamentals
Alkaline hydrolysis of ester is done with strong base {\\rm{NaOH}}NaOHand {\\rm{O}}{{\\rm{H}}^ - }OH\u2212acts as nucleophilic reagent. This reaction is reversible, since carboxylate anion has tendency to react with an alcohol and gives back ester.
Step-by-step
Step 1 of 2
Attack of [{ m{O}}{{ m{H}}^ - }]on carbonyl take place as follows resulting formation of tetrahedral intermediate:
\u043e\u043d\u043e\u043d
Explanation | Hint for next step
The {\\rm{O}}{{\\rm{H}}^ - }OH\u2212nucleophile attacks on the electrophilic carbon of an ester {\\rm{C}} = {\\rm{O}}C=O and forms tetrahedral intermediate after breaking the \\pi\u03c0-bond.
Step 2 of 2
Hydrolysis of product formed from step 1 followed by reaction with {\\rm{NaOH}}NaOHis as follows:
The products of the reaction are:
Explanation | Common mistakes
On the reaction of octyl acetate with aqueous sodium hydroxide, the products octyl alcohol and acetate ion are formed after omitting the {\\rm{N}}{{\\rm{a}}^ + }Na+ions.
To balance it, it would be N2 + 3H2 ------> 2NH3.
for c) it would be 2N2 + 6H2 -------> 4NH3