A(g) + B(g) rightarrow C(g) The rate law for the above reaction is: -d[A]/dt = k

Question

A(g) + B(g) rightarrow C(g) The rate law for the above reaction is: -d[A]/dt = k[A][B] The rate constant is 2.40 times 10-3 L mol-1 s-1 at 305.0 degree C and 3.41 L mol-1 s-1 at 512.0 degree C. Use the Arrhenius equation k = Ae-Ea/RT to: Calculate Ea for this reaction. 1.32 times 102 kJ/mol You are correct. Computer's answer now shown above. Your receipt is 163-244 ? Previous Tries Calculate the rate constant k at 382.0 degree C for this reaction. Tries 2/8 Previous Tries Calculate the pre-exponential factor A for this reaction.

Explanation / Answer

According to Arrherinus Equation log ( K' / K ) = ( Ea /2.303 R ) [ ( T' - T ) / T T' ]
From this Ea = [ ( 2.303 * R * T * T' ) / ( T' - T ) ] log ( K' / K) Given T = 305 oC = 305 + 273 = 578K T ' = 512 o C = 512 + 273 = 785 K K = 2.4 * 10^-3 L mol^-1 s^-1 K ' = 3.41 L mol ^ -1 s^-1 R = gas constant = 0.0821 L atm / mol - K Plug the values we have activation energy , E a = 1306.55 J / mol                                                                        =1.30655 KJ / mol                                                                        =1.30655 KJ / mol

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