Answer:
Explanation:
300 ml x300mm Hg/500 mm Hg= 180 ml
pressure goes up volume goes down
P1V1 =P2V2
V2= V1 x P1/P2
When: (CH3)3COOH(CH3)3⇒ C2H6 + 2CH3CoCH3
A B 2C
SO first for A:
When feed rate to reactor = FA(o), and Effluent rate from reactor= FA=Fa(o)(1-x)
Second for B:
When feed rate to reactor = 0, and Effluent rate from reactor FB= FA(o)X
Third for C:
When feed rate tp reactor = 0, and the Effluent rate from reactor FC= 2FA(o)X
So, the total feed rate to reactor FT(o) = FA(o)
and the total Effluent rate from reactor FT = FA(o)(1+2X)
For the CSTR:
V= FA(o) X / γa where -γ = KCA
We can get the total concentration at any point CT and the entrance CT(o) from this equation:
CT= FT/Q= P/ZRT
and CT(o) = FT(o)/ Q(o) = P(o)/ Z(o)RT(o)
We can neglect the changes in Z (the compressibility factor) So:
Q = Q(o) FT/FT(o) P(o)/P T/T(o)
Because there is no pressure drop and for the isothermal system we can assume P & T are constants:
∴ Q = Q(o) FT/ FT(o)
∴ CA = FA / Q = (FA(o)(1-X)) / (Q(o)(1+2X))
= CA(o)((1-X)/(1+2X)) -----> eq 1
When K (the reaction rate constant) at 127C° can be evaluated with R (the gas constant) 8.314 J/mole.K° = 0.08205 L.atm/mole.°K
When, K2=K1(exp)[E/R ( (1/T1) - (1/T2)]
when T1 by kelvin = 50+273= 323 K° and T2= 127+273 = 400 K°
and activition energy = 85 KJ/mol = 85000 J/mol
∴ K2 =10^-4 (exp) [85000/8314 ((1/323) - (1/400)] = 0.0443 min^-1
We can get the concentration of (A) at 10 atm and 127C° (400K°) by:
CA(o)= PA(o) /RT = 10 / ( 0.08205 * 400 ) = 0.305 mol / L
∴ The reactor volume of the CSTR is:
V = FA(o)X / KCA---> eq 2
by substitute eq 1 in eq 2
∴ V=[ FA(o)x * (1+2x)] / [ KCA(o) * (1-X) ]
= [ 2.5 * 0.9 * (1+1.8)] / [0.0443 * 0.305* 0.1]
= 4662.7 L
- The equilibrium conversion:
according to the reaction:
A⇔B + 2C
and we have:
-γ = KfCA - Kb CB( C(C)^2) = 0 and KC = Kf / Kb = (CB* CC^2)/ CA= 0.025(as it is agiven)
∴[ ( XCA(o) ) ( 2XCA(o)^2 ) ] / [ (1+2X)^2 * CA(o) * (1-X) ] = 0.025
when we have CA(o) (the initial concentration)= 0.305 mol/L, So we have this eq:
[(XCA(o)) (2XCA(o))^2] / [(1+2X)^2 CA(o) (1-X) ] = 0.025
By the matlab function solve
∴ X= X eq = 0.512 for 90% of the equilibrium conversion
∴ X= 0.9 X eq = 0.4608
Answer:
B. Conductivity.
Explanation:
Conductivity is the quantity of heat passing per second through a slab of unit cross-sectional area when the temperature gradient between the two faces is unity when put in heat.
Bitter taste (opposed to sour taste of acids)...
Slimy, or soapy feel on fingers (Slippery).....
Many bases react with acids and precipitate salts......
Strong bases may react violently with acids. An acid spill can be safely neutralised by using a mild base......
Bases turn red litmus paper blue.....
Bases are substances that contain metal oxides or hydroxides........
Bases which are soluble in water form alkalis (soluble bases)......
