Diffusion of Ammonia in an Aqueous Solution Ammonia (A)-water (B) solution ta 278 K and 4 mm thick is in contact with an organic
liquid at this interface. The concentration of ammonia in the organic phase is held constant and is such that the equilibrium concentration of ammonia at the interface is 2.0 wt% ammonia (density of aqueous solution is 991.7 Kg/m) The concentration of ammonia in water 4 mm away is 10 wt% (density of solution is 961.7 Kg/m). Water and the organic phase are insoluble in each other. The diffusion coefficient of NH3 in water is 1.24 × 10-9 m2/s.
(a) At steady state, calculate Na in Kmol/m-s
(b) Calculate NB. Explain.
(a) At steady state, calculate Na in Kmol/m-s
(b) Calculate NB. Explain.
1 answer:
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Answer:
5.21e-2mm
Explanation:
Please see attachment
Answer:
(a)
<em>d</em>Q = m<em>d</em>q
<em>d</em>q = <em>d</em>T
=
(T₂ - T₁)
From the above equations, the underlying assumption is that remains constant with change in temperature.
(b)
Given;
V = 2L
T₁ = 300 K
Q₁ = 16.73 KJ , Q₂ = 6.14 KJ
ΔT = 3.10 K , ΔT₂ = 3.10 K for calorimeter
Let be heat constant of calorimeter
Q₂ = ΔT
Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂
Q₁ - Q₂ = m ΔT
number of moles of n-C₆H₁₄, n = m/M
ρ = 650 kg/m³ at 300 K
M = 86.178 g/mol
m = ρv = 650 (2x10⁻³) = 1.3 kg
n = m/M => 1.3 / 0.086178 = 15.085 moles
Q₁ - Q₂ = m ' ΔT
= (16.73 - 6.14) / (15.085 x 3.10)
= 0.22646 KJ mol⁻¹ k⁻¹