Decomposition of A2(g) and B3(g) follow first-order kinetics as: A2(g) → 2A(g); K1 (/h^-1) = e^(-14000(J)/RT + 5) ; B3(g) → 3B(g); K2 (/h^-1) = e^(-20000(J)/RT + 10). One mole, each of A2(g) and B3(g)

Question

Decomposition of A2(g) and B3(g) follow first-order kinetics as: A2(g) → 2A(g); K1 (/h^-1) = e^(-14000(J)/RT + 5) ; B3(g) → 3B(g); K2 (/h^-1) = e^(-20000(J)/RT + 10). One mole, each of A2(g) and B3(g), is taken in a 100 l evacuated flask and at some temperature at which they start decomposing at the same rate. The incorrect information regarding the reactions is

Prashant Kotian · Accepted Answer

Correct answer: C,D. Equating rates yields K1=K2, which solves to T = 1200/R K. Because initial moles and rate constants match, PA2 always equals PB3. However, initial pressure is approx 0.23 atm and increases over time, so total pressure can never be less than 0.2 atm. The moles of A and B produced have a 2:3 ratio, so PA/PB = 2/3, not 1.0. Therefore, correct answer is C,D.

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