Balance each of the following examples of heterogeneous equilibria and write each kc expression. then calculate the value of kc for the reverse reaction.(1) al(s) + naoh(aq) + h2o(l) ⇋ na[al(oh)4](aq) + h2(g) kc for balanced reaction = 11(2) h2o(l) + so3(g) ⇋ h2so4 (aq) kc for balanced reaction = 0.0123 (3) p4(s) + o2(g) ⇋ p4o6(s) kc

(1) 2Al(s) + 2NaOH(aq) + 6H2O(l) ↔2Na[Al(OH)4](aq) + 3H2(g)

∴ Kc = ( PH2³ * [Na[Al(OH)4]² ) / [NaOH]² = 11

(2) H2O(l) + SO3(g) ↔ H2SO4(aq)

∴ Kc = [ H2SO4 ] / PSO3 = 0.0123

(3) 2P4(s) + 6O2(g) ↔ 2P4O6(s)

∴ Kc = Kc = 1 / PO2∧6

Explanation:

(1) 2Al(s) + 2NaOH(aq) + 6H2O(l) ↔ 2Na[Al(OH)4](aq) + 3H2(g)

∴ O /  Al: 0 → +2 ≡ 2e-

         Na: +1 → +2

∴ R /  H: +1 → 0

     2 - Al - 2

     2 - Na - 1

     8 - O - 8

     14 - H - 14

⇒ Kc = ( PH2³ * [Na[Al(OH)4]² ) / [NaOH]² = 11

(2) H2O(l) + SO3(g) ↔ H2SO4(aq)

           1 - S - 1

           4 - O - 4

           2 - H - 2

⇒ Kc = [ H2SO4 ] / PSO3 = 0.0123

(3) 2P4(s) + 6O2(g) ↔ 2P4O6(s)

       8 - P - 8

       12 - O - 12

⇒ Kc = 1 / PO2∧6

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