The rate of disappearance of the reactant X in the reversible reaction X ⇌ Y, at two temperatures, is given as -(d[X])/dt = (3.0×10 -3 s -1 )[X] - (1.0×10 -3 s -1 )[Y] (at 300 K) -(d[X])/dt = (6.0×10

Question

The rate of disappearance of the reactant X in the reversible reaction X ⇌ Y, at two temperatures, is given as

-(d[X])/dt = (3.0×10^-3 s^-1)[X] - (1.0×10^-3 s^-1)[Y] (at 300 K)

-(d[X])/dt = (6.0×10^-2 s^-1)[X] - (2.0×10^-3 s^-1)[Y] (at 400 K)

The enthalpy of reaction in this temperature range, in J/mol, is (R = 8.314 J/mol/K)

TestHub · Accepted Answer

At equilibrium -d[X]/dt = 0, so K = k f /k b At 300 K: K 1 = (3.0×10 -3 )/(1.0×10 -3 ) = 3 At 400 K: K 2 = (6.0×10 -2 )/(2.0×10 -3 ) = 30 log(K 2 /K 1 ) = log 10 = 1 Δ H = 2.303 Rlog(K 2 /K 1 )·(T 1 T 2 )/(T 2 - T 1 ) Δ H = 2.303×8.314×1×(300×400)/100 = 2.303×8.314×1200 Δ H = +2.30×10 4 J/mol This is option (a).

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