A chemist mixes 0.46mol of CO with 0.45mol of NO. The mixture is left to reach equilibrium at constant temperature. The student analyses the equilibrium mixture and finds that 0.25mol NO remains. The total volume of the equilibrium mixture is 1.0dm3.
I know how to calculate Kc but I can't figure out how to work out the concentrations I'm supposed to use
Please explain
Thank you
I know how to calculate Kc but I can't figure out how to work out the concentrations I'm supposed to use
Please explain
Thank you
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2CO + 2NO -> 2CO2 + N2
Initially
n(CO)=0.46mol n(NO)=0.45mol n(CO2)=0 n(N2)=0
If there if 0.25mol NO left at equilibrium then 0.20mol must have reacted. As CO and NO react in a one-to-one ratio 0.20mol of CO must have reacted too.
n(CO)eq= 0.46-0.20=0.26mol
For every mole of NO that reacts, one mole of CO2 is formed and half a mole of N2 is formed. As 0.20mol of NO reacts,
n(CO2)eq=0.20mol
n(N2)eq=0.10mol
Hence you work out Kc from
[NO]=0.25M [CO]=0.26M [CO2]=0.20M [N2]=0.10M
Initially
n(CO)=0.46mol n(NO)=0.45mol n(CO2)=0 n(N2)=0
If there if 0.25mol NO left at equilibrium then 0.20mol must have reacted. As CO and NO react in a one-to-one ratio 0.20mol of CO must have reacted too.
n(CO)eq= 0.46-0.20=0.26mol
For every mole of NO that reacts, one mole of CO2 is formed and half a mole of N2 is formed. As 0.20mol of NO reacts,
n(CO2)eq=0.20mol
n(N2)eq=0.10mol
Hence you work out Kc from
[NO]=0.25M [CO]=0.26M [CO2]=0.20M [N2]=0.10M
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Work it out from the volumes of the mixture and the amount of NO in it. Then subtract the concentration from 100 to find CO (they must both equal 100)