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 .
Part B question 1 Answer: C
that’s the on that makes most sense
Answer:
<em>Hello, Your answer will be </em><em>B) Jacob's backyard is on the north side of his house.</em>
<em>Hope That Helps!</em>
The answer for the following problem is mentioned below.
- <u><em>Therefore the final moles of the gas is 14.2 × </em></u>
<u><em> moles.</em></u>
Explanation:
Given:
Initial volume (
) = 230 ml
Final volume (
) = 860 ml
Initial moles (
) = 3.8 ×
moles
To find:
Final moles (
)
We know;
According to the ideal gas equation;
P × V = n × R × T
where;
P represents the pressure of the gas
V represents the volume of the gas
n represents the no of the moles of the gas
R represents the universal gas constant
T represents the temperature of the gas
So;
V ∝ n
= 
where,
(
) represents the initial volume of the gas
(
) represents the final volume of the gas
(
) represents the initial moles of the gas
(
) represents the final moles of the gas
Substituting the above values;
= 
= 14.2 ×
moles
<u><em>Therefore the final moles of the gas is 14.2 × </em></u>
<u><em> moles.</em></u>