At equilibirium,1.0 liter vessel was found to contain 0.2moles,0.2moles B,0.4 moles C,and 0.4 moles D.İf 0.1 moles A and 0.1mole
s B added to system what will be the new equilibirium concentration of A?
A+B=C+D
reversable reaction
A+B=C+D
reversable reaction
1 answer:
Answer:
See Explanation and Attachments => New concentration A = 0.10 mol/L
Explanation:
Explanation:
A + B ⇄ C + D
C(i): 0.20 mol 0.20 mol 0.4 mol 0.4 mol
ΔC: -0.10 mol -0.1 mol +0.1 mol +0.1 mol
C(eq): 0.10 mol 0.10 mol 0.5 mol 0.5 mol
__________________
=> A & B react to form more C & D as reaction shifts right to establish new equilibrium with new concentration values. See attached diagram. => I call this the 'chemical see-saw analysis'.
=> Also attached is a summary of stress factor effects of temperature and pressure-volume changes.
You might be interested in
Answer:
The volume of nitrogen oxide formed is 35.6L
Explanation:
The reaction of nitric acid with copper is:
Cu(s) + 4HNO₃ → Cu(NO₃)₂ + 2NO₂(g) + 2H₂O(l)
Moles of copper are:
Moles of nitric acid are:
As 1 mol of Cu reacts with 4 moles of HNO₃:
0.697 mol Cu × (4mol HNO₃ / 1mol Cu) = 2.79 moles of HNO₃ will react. That means Cu is limiting reactant.
Moles of NO₂ produced are:
0.697 mol Cu × (2mol NO₂ / 1mol Cu) = <em>1.394 moles of NO₂</em>
Using PV = nRT
<em>Where P is pressure (735torr / 760 = 0.967atm); n are moles (1.394mol); R is gas constant (0.082atmL/molK); T is temperature (28.2°C + 273.15 = 301.35K). </em>
Thus, volume is:
V = nRT / P
V = 1.394mol×0.082atmL/molK×301.35K / 0.967atm
V = 35.6L
<em>The volume of nitrogen oxide formed is 35.6L</em>