Answer to Question #234266 in Chemical Engineering for Lucky

Question

a) For a gas reaction at 400K, the rate is reported as -dPA/dt= 3.66 PA^2. What is the value of rate constant for this reaction if the rate equation is measured in terms of mol/ (l. h)?

Answer 1

First, we establish that if the rate is constant, then:

-r_A = -(1/V) Dn_A/dt = kC_A ² , mol/liter-hr

k = 3.66 = - (dP_A /dt) / pA² = (Atm/hr)² = (1/Atm)/ (Atm)²

We understand that:

pA=C_A  R T or dP_A = (RT) dC_A

With the provided figures, we determine the value of ‘k’ as:

- (RT) dC_A/dt = 3.66 (CA R T)²

= 3.66 (0.0820575) (400) = 120.1322

        We get: 120.1322 lit/mol-hr