at 22°C the equilibrium constant, Kc, for the following reaction is 4.66 x 10-3. (1) If 0.800 mol of N2O4 were injected into a closed 1.00-L container at 22°C, how many moles of each gas would be present at equilibrium? (b) If the volume were halved (to 0.500L) at constant temperature, how many moles of each gas would be present after the new equilibrium has been established? N2O4(g) 2NO2(g)

Question

at 22°C the equilibrium constant, Kc, for the following reaction is 4.66 x 10-3. (1) If 0.800 mol of N2O4 were injected into a closed 1.00-L container at 22°C, how many moles of each gas would be present at equilibrium?  (b) If the volume were halved (to 0.500L) at constant temperature, how many moles of each gas would be present after the new equilibrium has been established? N2O4(g)  2NO2(g)

DrBob222 · Answer

..........N2O4 ==> 2NO2
I........0.800.......0
C.........-x........2x
E........0.800-x....2x

Substitute the E line into Kc and calculate x and evaluate 0.800 -x. This give you M, the volume is 1 L, solve for mols.

b)
Calculate new (N2O4)= 0.800/0.5 = 1.600
Redo part a as above, x will be M, determine mols in 0.5L.