Consider the reaction: 2AB2 → A2 + 2B2 You start with one mole of AB2 and the reaction is done at constant pressure of 3.7 atm and 28.2 °C. Assume all gases behave like ideal gases. What is the value of the work (in J)?

-2503.2J

Explanation:

At constant pressure, change in work follows the equation:

W = -P*ΔV

Where W is work, P is pressure (3.7atm) and ΔV is change in volume

We can obtain change in volume using ideal gas low:

ΔV = ΔnRT/P

Where Δn is change in moles (3moles of products - 2 moles of reactants = 1)

R is gas constant (0.082atmL/molK)

T is absolute temperature (28.2°C + 273.15 = 301.35K)

P is pressure (3.7atm)

ΔV = ΔnRT/P

ΔV = 1mol*0.082atmL/molK*301.35K/3.7atm

ΔV = 6.69L

And work is:

W = -3.7atm*6.69L

W = -24.71atmL

1atmL = 101.3J

-24.71atmL * (101.3J / 1atmL) =

the electronic configuration of an atom with eight electrons is 1s²2s²2p⁴. the orbital diagrams are shown in the figure.

as per the hunds rule:

1. electrons tend to avoid being in the same orbital as per as practicable.

2. electrons in different orbitals of the same energy will have their spin parallel.

this can be illustrated from the multiplicity of the orbitals of the atom. as there are four electrons in the p orbital and p orbital has 3 sub-shells. the electrons can be fed in different ways as shown in the figure but the correct orbital diagram will be as per the hunds rule.