Answer is: because pure liquids (<span>shown in </span>chemical reactions<span> by appending (</span>l)<span> to the </span>chemical formula) and solids (<span>shown in </span>chemical equations by appending (s)<span> to the </span>chemical formula) not go in to he equilibrium constant expression, only gas state (shown in chemical reactions by appending (g) to the chemical formula) reactants and products go in to he equilibrium constant expression.
For example, equilibrium constant expression Kp for reaction:
A(s) + 2B(s) ⇄ 4C(g) + D(g).<span>
will be: Kp = [C]</span>⁴<span>·[D].
But for reaction </span>A(g) + 2B(g) ⇄ 4C(g) + D(g), will be:<span>
Kp = [C]</span>⁴<span>·[D] / [A]·[B]².</span>
Answer:
NA = 6.8 E-12 Kg H2(g) / hour
Explanation:
steady-state diffusion of A through non-diffuser B:
- NA = (DAB/RTz)(p*A1 - p*A2)
∴ (A): H2(g)
∴ (B): Pd
∴ DAB = 1.7 E-8 m²/s
∴ p*A1 = 2.0 Kg H2 / m³ Pd
∴ p*A2 = 0.4 Kg H2 / m³ Pd
∴ z = 6 mm = 6 E-3 m
∴ T = 600°C ≅ 873 K
∴ R = 8.314 J/mol.K = 8.314 N.m/mol.K
⇒ NA = ((1.7 E-8)/(8.314)(873)(6 E-3))(2.0 - 0.4)
⇒ NA = 6.246 E-10 mol/s.m³
for A = 0.25 m²
⇒ volume (v) = A×z = (0.25)(6 E-3) = 1.5 E-3 m³
∴ Mw H2(g) = 2.016 g/mol
⇒ NA = (6.246 E-10 mol/s.m³)(1.5 E-3 m³)(2.016 g/mol)(Kg/1000 g)(3600 s/h)
⇒ NA = 6.8 E-12 Kg H2(g)/h
Well when you would "get rid of" the water it would eventually be led to some ocean, pond, lake, etc. There it would eventually evaporate, rising up to the sky. This is where it would move around with clouds until it rained again.
