Answer:
a) pH = 4.213
b) % dis = 2 %
Explanation:
Ch3COONa → CH3COO- + Na+
CH3COOH ↔ CH3COO- + H3O+
∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]
mass balance:
⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [ CH3COOH ] + [ CH3COO- ]
<em>∴ C </em>CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)
∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL
⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]
charge balance:
⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water
⇒ [ CH3COO- ] = [ H3O+ ] + 1.00
⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5
⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]
⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0
⇒ [ H3O+ ] = 6.12 E-5 M
⇒ pH = - Log [ H3O+ ] = 4.213
b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4
∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol
⇒ % dis = 3.4 / 1.7 = 2 %
Answer:
3224 kJ/mol
Explanation:
The combustion of benzoic acid occurs as follows:
C₇H₆O₂ + 13/2O₂ → 7CO₂ + 3H₂O + dE
The change in temperature in the reaction is the change due the energy released, that is:
3.256K * (10.134kJ / K) = 33.00kJ are released when 1.250g reacts
To find the heat released per mole we have to find the moles of benzoic acid:
<em>Moles benzoic acid -Molar mass: 122.12g/mol-:</em>
1.250g * (1mol / 122.12g) = 0.0102 moles
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The dE combustion per mole of benzoic acid is:
33.00kJ / 0.0102moles =
<em>3224 kJ/mol </em>
Answer:
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The rate law equation is different from the determination of the equilibrium constant keq expression because rate law equation is used experimentally
And by looking at the equation of the reaction and chemicals used in the reaction equation Keq is determined.
<span>The value of the reaction quotient, when the chemical reaction approaches the state of equilibrium, this is called equilibrium constant.</span>
