Chemical Reaction Engineering homework

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SUMMARY

The decomposition of phosphine (4PH3(g) + P4(g) + 6H2(g)) is first-order with a rate constant k = 0.0155 s-1 at 953 K. For a 40% conversion of PH3, the time required for the reaction in a constant-volume batch reactor is calculated to be 32.96 seconds. Additionally, the mole fraction of H2 in the reaction mixture can be determined based on the stoichiometry of the reaction, where each mole of reacted PH3 produces 1.5 moles of H2.

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herbertpbarreto
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Homework Statement


The decomposition of phosphine 4PH3(g) + P4(g) + 6H2(g) is first-order with respect to phosphine with a rate constant k = 0.0155 s-l at 953 K. If the decomposition occurs in a constant-volume batch reactor at 953 K, calculate, for 40% conversion of PH3,

(a) the time required, s;
(b) the mole fraction of H2 in the reaction mixture.

Homework Equations


(
-rA)=-dCa/dT
(-rA)=kCa
Ca= Ca0 (1-Xa)

The Attempt at a Solution



a) i attempet do apply the equation for spatial time I am batch reactors:
-dCa/dT = kCa
∫-dCa/Ca =∫kdT
-ln(Ca/Ca0) = kT
-ln (Ca0(1-Xa)/Ca0) = kT
-ln (1-Xa) = kT --> -ln (1-0,4) = 0,0155 * T --> -ln(0,6)/0,0155 =T
T = 32,96 s
b) i have no idea
 
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herbertpbarreto said:

Homework Statement


The decomposition of phosphine 4PH3(g) + P4(g) + 6H2(g) is first-order with respect to phosphine with a rate constant k = 0.0155 s-l at 953 K. If the decomposition occurs in a constant-volume batch reactor at 953 K, calculate, for 40% conversion of PH3,

(a) the time required, s;
(b) the mole fraction of H2 in the reaction mixture.

b) i have no idea

So of the total original molecules of PH3 0.6 of them are still PH3, and of the 0.4 that have reacted each one has become 1.5 molecules of H2 and ¼ of P4 according to your formula (which should have a → instead of a + ) and you have any difficulty in finding what fraction H2 molecules are of the total? :))
 
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