[Frederick95]

At 35 degrees Celsius, 2.0 mol of pure NOCL(g) is introduced into a 2.0 L flask. The NOCL(g) partially decomposes according to the following equilibrium equation: 2NOCL(g)   ---> 2NO(g) + Cl2(g)

At equilibrium, the concentration of NO(g) is 0.032 mol/L. Use an ICE table to determine equilibrium concentrations of NOCL(g) and Cl2(g) at this temperature.

After making my ICE Table, I did the following calculation

2.0 - 2x = 0.032
        2x = 2.0 - 0.032
          x = 0.984
Since x = 0.984, then [NOCl(g)] = 0.984 mol/L; [ and Cl2(g)] = 1.986 mol/L


Can anyone help me get the correct answer, which is actually

[NOCl(g)] = 0.968 mol/L; [Cl2(g)] = 0.016 mol/L?

[boostar]

2.0 - 2x = 0.032
        2x = 2.0 - 0.032
          x = 0.984
Since x = 0.984, then [NOCl(g)] = 0.984 mol/L; [ and Cl2(g)] = 1.986 mol/L

Are you sure you made your ICE table? Post your ICE table here, then we can see where you have made mistakes. Once again, just like in your other thread:

At 35 degrees Celsius, 2.0 mol of pure NOCL(g) is introduced into a 2.0 L flask. The NOCL(g) partially decomposes according to the following equilibrium equation: 2NOCL(g)   ---> 2NO(g) + Cl2(g)

At equilibrium, the concentration of NO(g) is 0.032 mol/L. Use an ICE table to determine equilibrium concentrations of NOCL(g) and Cl2(g) at this temperature.

[Frederick95]

My ICE Table:

     2NOCl(g)  ----> 2NO(g)  +  Cl2(g)
I     2.0                     0            0
C    -2x                  +2x           x
E     2.0 - 2x             2x           x

[enahs]

When solving, are you raising them to the appropriate power?

I.E., in the Ice Table it is "2x", but when solving mathematically it is (2x)².

No?

[Frederick95]

No that doesnt work out

[enahs]

Yes, because you are making this much harder then you have to.

You have the equation:

2NOCl_(g)   ---> 2NO_(g) + Cl_2(g)

You are also told at Equilibrium you have 0.032 Mol/L of NO. You get twice as much NO gas as you do Cl₂ gas in this reaction, so therefor you must have 0.032/2 = 0.016 mols/L of Cl₂ gas. Yes?

But you also know 2 NOCl turned into 2 NO, or in other words, for every mol of NO produced a mol of NOCl was used up, yes? So if you started with 1 Mol/L of NOCl, and convert 0.032 Mol/L into NO gas, how much is left?


Make the math simpler for you. You started with 2 mols in 2 L.
2 mol/ 2L = 1 mol/ 1L