Check My Chemical Equilibrium Problem! 4.2 mol of oxygen and 4.0 mol of NO are introduced to an evacuated 0.50 L reaction vessel. At a specific temperature, the equilibrium 2NO(g) + O2(g) 2NO2(g) is reached when [NO] = 1.6 M. Calculate Kc for the reaction at this temperature? I did this: 1) NO: (4.0/ .5L)= 8M O2: (4.2/ .5L)= 8.4M NO= 1.6M 2) 2NO +O2 2NO2 I 8 8.4 0 C 6.4 3.2 6.4 E 1.6 5.2 6.4 3) [6.4]²/ [1.6]²[5.2] = 3.1M Is this right?? Let me know if anyone can! Please and thank you :D

Question

Check My Chemical Equilibrium Problem! 4.2 mol of oxygen and 4.0 mol of NO are introduced to an evacuated 0.50 L reaction vessel. At a specific temperature, the equilibrium 2NO(g) + O2(g) <-----> 2NO2(g) is reached when [NO] = 1.6 M. Calculate Kc for the reaction at this temperature? I did this: 1) NO: (4.0/ .5L)= 8M O2: (4.2/ .5L)= 8.4M NO= 1.6M 2) 2NO +O2<-----> 2NO2 I 8 8.4 0 C 6.4 3.2 6.4 E 1.6 5.2 6.4 3) [6.4]²/ [1.6]²[5.2] = 3.1M Is this right?? Let me know if anyone can! Please and thank you :D

aaronq · Answer

it looks right.

anonymous · Answer

Thanks!! Just want to make sure!

.sam. · Answer

From here, shouldn't it be
          2NO   +     O2<-----> 2NO2
I           8             8.4                    0 
C         -2x          -x                  +2x
E        8-2x       8.4-x             2x=1.6
           6.4           7.6                 0.8

$$K_c=\frac{0.8^2}{6.4^2(7.6)}$$

aaronq · Answer

you're mistaken, at eq, [NO]=1.6, not [NO2].