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 .
Answer:24.31
Explanation:Contribution made by isotope of mass 23.99= 23.99×78.99=1894.97
Contribution made by isotope of mass 24.99=24.99×10.00=249.9
Contribution made by isotope of mass 25.98=25.98×11.01=286.04
Total contribution=1894.97+249.9+286.04=2430.91
Average mass=2430.91÷100
=24.31
The total number of elements that one particular element can bond to can be determined by simply drawing the Lewis structure of the element.
Place the chemical symbol
Then look at the group number = valence electrons
Distribute the valence electrons around the atom.
C = 4 bonds
N = 3 bonds
O = 2 bonds
To determine what gas is this, we use Graham's Law of Effusion where it relates the rates of effusion of gases and their molar masses. We do as follows:
r1/r2 = √(M2 / M1)
Let 1 be the the unkown gas and 2 the H2 gas.
r1/r2 = 0.225
M2 = 2.02 g/mol
0.225 = √(2.02 / M1)
M1 = 39.90 g/mol
From the periodic table of elements, most likely, the gas is argon.
