Consider this reaction: KOH + HBr --> KBr + H2O
2 answers:
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Answer:
The rate of the reaction will increase by a factor of 9.
Explanation:
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In this case, considering the given second-order reaction, whose rate law results:
We easily infer that at constant concentration of A but tripling the concentration of B, we are going to obtain the following increasing factor while holding the remaining variables constant:
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(1) MO₂(s) + C(s) → M(s) + CO₂ (g), ΔG₁ = 288.9 kJ/mol
(2) C(s) + O₂(g) → CO₂(g), ΔG₂ = -394.4 kJ/mol
By adding both equations 1 + 2 we get the coupled reaction:
MO₂(s) + 2 C(s) + O₂(g) → M(s) + 2 CO₂(g)
ΔG⁰ = ΔG₁ + ΔG₂
= 288.9 + (-394.4) = -105.5 kJ/mol = -105500 J/mol
Temperature T = 25 + 273.15 = 298.15 K
Molar gas constant R = 8.314 J/mol.K
K =
=
= 3.05 x 10¹⁸
(2) C(s) + O₂(g) → CO₂(g), ΔG₂ = -394.4 kJ/mol
By adding both equations 1 + 2 we get the coupled reaction:
MO₂(s) + 2 C(s) + O₂(g) → M(s) + 2 CO₂(g)
ΔG⁰ = ΔG₁ + ΔG₂
= 288.9 + (-394.4) = -105.5 kJ/mol = -105500 J/mol
Temperature T = 25 + 273.15 = 298.15 K
Molar gas constant R = 8.314 J/mol.K
K =
=
= 3.05 x 10¹⁸