Question #234266

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)?

Expert's answer

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) / pA^{2 }= (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**

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