For this heterogeneous system 2 A ( aq ) + 3 B ( g ) + C ( l ) − ⇀ ↽ − 2 D ( s ) + 3 E ( g ) 2A(aq)+3B(g)+C(l)↽−−⇀2D(s)+3E(g) the concentrations and pressures at equilibrium are [ A ] = 5.66 × 10 − 2 M [A]=5.66×10−2 M , P B = 5.69 × 10 3 Pa PB=5.69×103 Pa , [ C ] = 9.14 M [C]=9.14 M , [ D ] = 15.78 M [D]=15.78 M , and P E = 9.33 × 10 4 torr PE=9.33×104 torr . Calculate the thermodynamic equilibrium constant, K K .

K = 1.08 X 10¹²

Explanation:

The solved solution is in the attach document

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol