Consider the reaction below: A(sol) ? B(sol) A series of experiments using a 1.000 M solution of A was heated at different temperatures. After some time, the data below were obtained. Answer the following questions: Temperature (°C) k(s?¹) 145.00 9.295e ? 08 155.83 1.373e ? 07 180.75 3.371e ? 07 196.00 5.840e ? 07 a) What is the activation energy (E?) for this reaction? E?(kJ) = number (rtol=0.2, atol=1e-08) b) Determine the pre-exponential factor A? A = number (rtol=0.2, atol=1e-08) c) At what temperature is the rate constant equal to 1? T(°C) = number (rtol=0.2, atol=1e-08)
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314 J/mol*K), and T is the temperature in Kelvin. Taking the natural logarithm of both sides, we get: ln(k) = ln(A) - (Ea/RT) We can plot ln(k) vs. 1/T and use the slope of the line to find -Ea/R. Then, we can solve for Ea: slope = -Ea/R Ea = -slope * R Using Show more…
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The following questions relate activation energy, Ea, and temperature. a) To determine the activation energy for a reaction, the variation of reaction rate with temperature is investigated. The relationship of the rate constant, k, for the reaction is related to the absolute temperature using the Arrhenius equation, a linear version of which is detailed below: ln k = -Ea/R x (1/T) + constant where R is the gas constant, 8.314 J mol-1 K-1 i) State what specific graph could be used to obtain the activation energy for the reaction. You should identify the x- and y-axes. (1) ii) When such a graph was plotted, the gradient of the graph was found to be -4029.3. Calculate the activation energy for this reaction, in kJ mol-1. (1) b) Consider a reaction involving gases, which has an activation energy of +75 kJ mol-1. Calculate the fraction of collisions that will have sufficient energy to react at a temperature of 323 K. (3)
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