Question

Calculate the change in the number of moles of each gas from the start of a reaction until reaching equilibrium?

The following equilibrium reaction takes place at 177°C.
2 NO(g) + O2(g) ⇌ 2 NO2(g)
If the amount of (NO2) gas is exactly 0 mol at the start of the reaction and
2.19 ✕ 10−4 mol when the reaction reaches equilibrium,
Calculate the change in the number of moles of each gas from the start of the reaction to the equilibrium stage.
Within this answer, A positive sign (+) indicates an increase in an amount; a negative sign (-)indicates a decrease in an amount.

The change in the number of moles of NO2 =
The change in the number of moles of NO =
The change in the number of moles of O2 =

I honestly have no idea where to start here. Please help with a teaching explanation. Thank you.

Answer 1

Use a R.I.C.E. diagram

Explanation:

R stands for Reaction
I stands for initial concentration
C stands for change in concentration
E stands for equilibrium concentration

`"R":color(white)(mmmm)"2NO(g)" + "O"_2"(g)" ⇌ "2NO"_2"(g)"`
`"I/mol·L"^"-1":color(white)(mmmmmmmmmmmll)0"`
`"C/mol·L"^"-1": color(white)(m)"-2"xcolor(white)(mmm)"-"xcolor(white)(mmmm)"+2x"`
`"E/mol·L"^"-1":color(white)(mmmmmmmmm)2.19 xx 10^"−4"`

The change in concentrations were derived from the coefficients in the reaction

Although I can't complete this RICE diagram, because the initial concentrations of reactants are not given, I can still figure out the change in mols even for the reactants
`2x=2.19 ✕ 10^(−4)`
`x=1.095*10^(-4) mol`
So the change for each
NO₂= `2.19 ✕ 10^(−4) mol`
NO= `-2.19 ✕ 10^(−4) mol`
O₂= `-1.095*10^(-4) mol`