Asked by Anonymous
Suppose that 2.00 moles of HCl in a 1.00L glass flask slowly decomposes into H2 and Cl2. When equilibrim is reached, the concentrations of H2 and Cl2 are both 0.214M. What is the Keq?
Answers
Answered by
unknown
Keq = product/ reactant
Write the balanced equation for what you are given. 2HCl--->H2 + Cl2
I believe that's how you start off.
Write the balanced equation for what you are given. 2HCl--->H2 + Cl2
I believe that's how you start off.
Answered by
DrBob222
.........2HCl ==> H2 + Cl2
I........2M........0.....0
C........-2x.......x.....x
E.......2-2x.....0.214..0.214
Therefore, at equilibrium, H2 = 0.214, Cl2 = 0.214 and HCl = 2-2*0.214. Substitute into Keq expression and evaluate for Keq.
I........2M........0.....0
C........-2x.......x.....x
E.......2-2x.....0.214..0.214
Therefore, at equilibrium, H2 = 0.214, Cl2 = 0.214 and HCl = 2-2*0.214. Substitute into Keq expression and evaluate for Keq.
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