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.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol