PH of a solution will be <span>higher than 7
</span>
Ammonium cyanide is a salt formed by hydrogen cyanide and ammonia. Ammonia is a weak base and hydrogen cyanide is a weak acid.
NH₄CN + H₂O ⇒ NH₃ + HCN
NH₄⁺ + H₂O -----> H₃O⁺ + NH₃
CN⁻ + H₂O -----> HCN + OH⁻
Although both compounds are weak electrolytes, NH₃ is somewhat stronger base than HCN is a strong acid, so the solution reacts alkaline. We can prove this using Ka and Kb values:
Ka(HCN) = 4.9 x × 10⁻¹⁰
Kb(NH₃) = 1.8 × 10⁻⁵<span>
Kw= </span>1.0 × 10⁻¹⁴
Let's first calculate Ka for NH₄⁺:
Ka(NH₄⁺) x Kb(NH₃<span>) = pKw
</span>Ka(NH₄⁺) = Kw/Kb(NH₃) = 5.6 x 10⁻¹⁰
Then, Kb for CN⁻:
Kb(CN⁻) x Ka(HCN) = pKw
Kb(CN⁻) = Kw/Ka(HCN) = 2 x 10⁻⁵
From this, we can see that the acid constant NH4⁺ is much lower than the base constant of CN⁻, which will say that the solution of NH₄CN will react slightly alkaline because of the higher presence of hydroxyl ions in solution.
B.
And maybe put your question in the English/Literature tag next time lol
Answer:
The answer to your question is: ΔH = -283 kJ/mol, first option
Explanation:
Reaction
CO + O₂ ⇒ CO₂
ΔH = ∑H products - ∑H products
ΔH = -393.5 - (-110.5 + 0)
ΔH = -393.5 + 110.5
ΔH = -283 kJ/mol
