Use standard enthalpies of formation to determine for: 2al(s) + 3cl2(g) → 2alcl3(s) enter in kj. find the change in internal energy for this reaction. enter in kj.

Hey there!:

Al= 0 Kj

Cl2 = 0 kj

Aluminium Chloride Solid AlCl3  = 705.63 KJ

Therefore:

enthalpy of formation is :

2*(-705.63 KJ) = 1411.26 KJ

Hope that helps!

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

the answer is 0.8m.

by: nicholas