Answer:
The solution to this question is 5.153×10⁻⁴(kmol)/(m²·s)
That is the rate of diffusion of ammonia through the layer is
5.153×10⁻⁴(kmol)/(m²·s)
Explanation:
The diffusion through a stagnant layer is given by
Where
= Diffusion coefficient or diffusivity
z = Thickness in layer of transfer
R = universal gas constant
= Pressure at first boundary
= Pressure at the destination boundary
T = System temperature
= System pressure
Where = 101.3 kPa 
, 
, 
0.5×101.3 = 50.65 kPa
Δz = z₂ - z₁ = 1 mm = 1 × 10⁻³ m
R = 
T = 298 K and = 1.18 
= 1.8×10⁻⁵
= 5.153×10⁻⁴
Hence the rate of diffusion of ammonia through the layer is
5.153×10⁻⁴(kmol)/(m²·s)
Answer:
Explanation:
a) for shifting reactions,
Kps = ph2 pco2/pcoph20
=[h2] [co2]/[co] [h2o]
h2 + co2 + h2O + co + c3H8 = 1
it implies that
H2 + 0.09 + H2O + 0.08 + 0.05 = 1
solving the system of equation yields
H2 = 0.5308,
H2O = 0.2942
B) according to Le chatelain's principle for a slightly exothermic reaction, an increase in temperature favors the reverse reaction producing less hydrogen. As a result, concentration of hydrogen in the reformation decreases with an increasing temperature.
c) to calculate the maximum hydrogen yield , both reaction must be complete
C3H8 + 3H2O ⇒ 3CO + 7H2( REFORMING)
CO + H2O ⇒ CO2 + H2 ( SHIFTING)
C3H8 + 6H2O ⇒ 3CO2 + 10 H2 ( OVER ALL)
SO,
Maximum hydrogen yield
= 10mol h2/3 molco2 + 10molh2
= 0.77
⇒ 77%
Water vapor and carbon dioxide!
Answer:
Given:
high temperature reservoir 
low temperature reservoir 
thermal efficiency 
The engines are said to operate on Carnot cycle which is totally reversible.
To find the intermediate temperature between the two engines, The thermal efficiency of the first heat engine can be defined as
The thermal efficiency of second heat engine can be written as
The temperature of intermediate reservoir can be defined as
