For this heterogeneous system 2A(aq)+3B(g)+C(l)↽−−⇀2D(s)+3E(g) 2A(aq)+3B(g)+C(l)↽−−⇀2D(s)+3E(g)
the concentrations and pressures at equilibrium are [A]=2.25×10−2 M[A]=2.25×10−2 M,
PB=1.09×103 PaPB=1.09×103 Pa.
[C]=13.08 M[C]=13.08 M.
[D]=17.00 M[D]=17.00 M.
PE=3.20×104 torrPE=3.20×104 torr. Calculate the thermodynamic equilibrium constant, K.

Explanation:

Hello,

In this case, for the reaction:

We notice it is undergone as heterogeneous equilibrium, that is why, for the thermodynamic equilibrium constant we consider the aqueous species only, that is the concentration of A to the second power (stoichiometric coefficient) only:

Thus, the equilibrium constant turns out:

Clearly, B, C, D and E are not considered as they are gaseous, liquid, solid and gaseous respectively, and based on the procedure to write the law of mass action, they are not considered due to different phases.

Best regards.

c-14 is carbon-14, or radiocarbon.   it is a radioactive isotope of carbon with an atomic nucleus containing 6 protons and 8 neutrons.