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KengaRu [80]
3 years ago
10

The ka of phosphoric acid, h3po4, is 7.6  10–3 at 25 °c. for the reaction h3po4(aq) h2po4 – (aq) + h+ (aq) ∆h° = –14.2 kj/mol.

what is the ka of h3po4 at 60 °c?
Chemistry
1 answer:
blsea [12.9K]3 years ago
7 0
<span>Use the van't Hoff equation: ln ( K2 K1 ) = Δ Hº R ( 1 T1 ⒠1 T2 ) ln ( K2 7.6*10^-3 ) = -14,200 J 8.314 ( 1 298 ⒠1 333 ) ln ( K2 7.6*10^-3 ) = ⒠1708 ( 0.00035 ) ln ( K2 0.0076 ) = ⒠0.598 Apply log rule a = log b b a -0.598 = ln ( e ⒠0.598 ) = ln ( 1 e 0.598 ) Multiply both sides with e^0.598 K 2 e 0.598 = 0.0076 K e 0.598 e 0.598 = 0.0076 e 0.598 K 2 = 0.0076 e 0.598 = 4.2 ⋅ 10 ⒠3 K2 = 4.2 ⋅ 10 ⒠3</span>
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Ostrovityanka [42]

Explanation:

The given data is as follows.

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                            = 286000 J

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 S_{O_{2}} = 205 J/^{o}K

Hence, formula to calculate entropy change of the reaction is as follows.

          \Delta S_{rxn} = \sum \nu_{i}S_{i}_(products) - \sum \nu_{i}S_{i}_(reactants)

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                    = 163.5 J/K

Therefore, formula to calculate electric work energy required is as follows.

             \Delta G_{rxn} = \Delta H_{rxn} - T \Delta S_{rxn}

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                            = 237.277 kJ

Thus, we can conclude that the electrical work required for given situation is 237.277 kJ.

6 0
3 years ago
Determine the molar solubility of pbso4 in pure water. ksp (pbso4) = 1.82 x 10-8.
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Use the ICE table approach as solution:

           PbSO₄   --> Pb²⁺ + SO₄²⁻
I             -                 0          0
C           -                +s         +s
E           -                  s          s

Ksp = [Pb²⁺][SO₄²⁻]
1.82×10⁻⁸ = s²
Solving for s,
s = <em>1.35×10⁻⁴ M</em>
5 0
3 years ago
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