NEED HELP WITH CHEMISTRY EQUILIBRIUM PROBLEM

Question

a. Vanillic acid reacts in water as follows: HC8H7O4(aq) + H2O (l) -----> H3O+(aq) + C8H7O4^–(aq) Kc= 3.1 x 10^-5 Determine the equilibrium concentrations of HC8H7O4, H3O+ and C8H7O4^–. Start by setting up an ICE table. The initail concenration of HC9H7O4 is 0.00714M while the concentrations of H3O+and C8H7O4^–are initially zero (0 M). Complete the table by expressing changes using a variable, x, and incorporate this variable into the listed equilibrium concentration.

b. Consider the reaction: 

Ni (s) + 4CO (g) --->Ni(CO)4(g) Kc= 375 at 50 oC. The conditions in a sealed reaction vessel are as follows: the concentrations of CO and Ni(CO)4 are 0.181 M and 0.526 M respectively. There is also some Ni in the flask. Does the concentration of Ni(CO)4 increase, decrease, or remain the same over time? In other words, do the specified conditions correspond to equilibrium conditions at this temperature? If not, in which direction does the reaction need to move to achieve equilibrium?

Answer 1

HC8H7O4(aq) + H2O (l) -----> H3O+(aq) + C8H7O4^–(aq) 

0.00714......................................0...................0.........Initial

-x...............................................+x.................+x.......Change

0.00714-x..................................x....................x........Equilibrium

3.1x10^-5 = (x)(x)/0.00714 - x)

Solve for x to get the equilibrium concentrations of both H3O^+ and C8H7)4^-. The equilibrium concentration of HC8H7O4 will be 0.00714 - x.

b) Ni(CO)4 concentration will decrease. Because Ni(s) indicates it is a solid, it will not appear in the Keq/Kc equation. Find the value of Q: Q = [Ni(CO)4/[CO[^4 = 0.526/0.00107 = 491

Since Q > Kc, the reaction must shift to the left to achieve equilibrium.