5.80 mol of solid a was placed in a sealed 1.00-l container and allowed to decompose into gaseous b and c. the concentration of b steadily increased until it reached 1.20 m, where it remained constant.
a(s) equilibrium b(g) +c(g)
then, the container volume was doubled and equilibrium was re-established. how many moles of a remain?
hint: start by finding the value of k. make a table that expresses the initial and final amounts of each species. we can use moles in the table so long as we convert back to concentrations before plugging into the k expression. fill in the missing values in this table, then calculate kc. hint: you can use the stoichiometry of the reaction to determine the changes in a and c.

3.40 mol

Explanation:

By the reaction equation, the number of moles of B and C is equal, so their concentration must be equal. To determinate the equilibrium constant, the solids are not put in the expression, because their activity is equal to 1, so:

Kc = [B]¹x[C]¹

Kc = 1.2 x 1.2 = 1.44

By the Le Chatilers principle, a change in the volume will change the pressure in the system and will shift the equilibrium. But Kc only changes with the temperature, so the concentration of B and C remains 1.2 mol/L. To find the number of moles, we multiply the concentration by the new volume (2.00 L)

nB = nC = 1.2 x 2 = 2.4 mol

So, because the stoichiometry is 1 mol of A: 1 mol of B: 1 mol of C, it was consumed 2.4 mol of A, then remains:

nA = 5.80 - 2.40 = 3.40 mol

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"