Temperature of 62.0 °C is placed in 50.0 mL of H2O initially at 25.0 °C and allowed to reach thermal equilibrium
Answer:
Products are favored.
Explanation:
The acid-base reaction of CH₃COOH (acid) with NH₃ (base) produce:
CH₃COOH + NH₃ ⇄ CH₃COO⁻ + NH₄⁺ Kr = ?
It is possible to know Kr of the reaction by the sum of acidic dissociations of the half-reactions. That is:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ Ka = 1.8x10⁻⁵
NH₃ + H⁺ ⇄ NH₄⁺ 1/Ka = 1/ 5.6x10⁻¹⁰ = 1.8x10⁹
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CH₃COOH + NH₃ ⇄ CH₃COO⁻ + NH₄⁺ Kr = 1.8x10⁻⁵×1.8x10⁹ = <em>3.2x10⁴</em>
<em> </em>
As Kr is defined as:
Kr = [CH₃COO⁻] [NH₄⁺] / [CH₃COOH] [NH₃]
And Kr is > 1
[CH₃COO⁻] [NH₄⁺] > [CH₃COOH] [NH₃],
showing <em>products are favored</em>.
Answer:
190 °C
Step-by-step explanation:
The pressure is constant, so this looks like a case where we can use <em>Charles’ Law</em>:
V₁/T₁ = V₂/T₂ Invert both sides of the equation.
T₁/V₁ = T₂/V₂ Multiply each side by V₂
T₂ = T₁ × V₂/V₁
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V₁ = 3.75 L; T₁ = (37 + 273.15) K = 310.15 K
V₂ = 5.6 L; T₂ = ?
=====
T₂ = 310.15 × 5.6/3.75
T₂ = 310.15 × 1.49
T₂ = 463 K
t₂ = 463 – 273.15
t₂ = 190 °C
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