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.

1 Answer
Hriman
Feb 24, 2018

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#

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