Question
Can someone explain in detail how to do this problem? The answer is 240 kJ/mol, but I keep getting around 120kJ/mol. Since the real answer is twice what I am getting, I mightonly be making a minor error, but I can't figure it out. The second order rate constant for the decomposition ofnitrous oxide into nitrogen molecule and oxygen atom has beenmeasured at different temperatures. k(1/M xs) t(oC) 1.87 x10-3 600 0.0113 650 0.0569 700 0.244 750 Determine graphically the activation energy for thereaction. Can someone explain in detail how to do this problem? The answer is 240 kJ/mol, but I keep getting around 120kJ/mol. Since the real answer is twice what I am getting, I mightonly be making a minor error, but I can't figure it out. The second order rate constant for the decomposition ofnitrous oxide into nitrogen molecule and oxygen atom has beenmeasured at different temperatures. k(1/M xs) t(oC) 1.87 x10-3 600 0.0113 650 0.0569 700 0.244 750 Determine graphically the activation energy for thereaction. 0.0113 650 0.0569 700 0.244 750 Determine graphically the activation energy for thereaction.
Explanation / Answer
We Know that : According to Arrehenius Equation: log K2/ K1 = Ea / 2.303R ( 1 / T1 - 1 / T2 ) log [ 0.0113/ 1.87 x 10-3 ] = Ea / 2.303 x 8.314 J / mol- K x [ 1 / 873 - 1 / 923 ] K Ea = 241357 J / mol = 241.357 KJ / mol The abovevalue can also be obtained by plotting the graph From the graph we have to obtain the slope of the line . slope = - Ea / R