The rate constant for a reaction is measured as a function of time. A plot is created by graphing ln(k) on the y-axis and 1/T on the x-axis, and a best fit line with a slope of -83,871 K is obtained. Based on this data what is the activation energy of this reaction (in kJ/mol)?
R= 8.314 J/(K mol)
The rate constant for a reaction is measured as a function of time. A plot is created...
The activation energy for a particular reaction is 102 kJ/mol. If the rate constant is 35 × 10⁻⁴ s⁻¹ at 322 K, what is the rate constant at 273 K? An experimental plot of ln(k) vs. 1/T is obtained in lab for a reaction. The slope of the best-fit line for the graph is -3090 K. What is the value of the activation energy for the reaction in kJ/mol?
The rate constant (k) for a reaction was measured as a function of temperature. A plot of ln k versus 1/T(ln k) is linear and has a slope of -5375 K. Calculate the activation energy for the reaction.
The equilibrium constant K for a certain reaction was measured as a function of temperature (in Kelvin). A plot of ln(K) versus 1/T for this reaction gives a straight line with a slope of 1.352 times 10^4 K and a y-intercept of -14.51. Determine the value of Delta H degree for this reaction -120.6 kJ/mol -164.7 kJ/mol -112.4 kJ/mol -176.7 kJ/mol none of the above
The rate of a reaction was measured as a function of time and
the data was plotted as shown in the figure below. The slope of the
plot is -5.17 × 10 K. The y-intercept is -1.69. What is the
activation energy of the reaction?
The answer is 43.0 kj/mol, probably rounded
In-k
the rate constant (k) for a reaction was measured as a function of temperature. A plot of lnk versus 1/T(in K) is linear and has a slope of -9.90•10^3 K. calculate the activation energy for the reaction.
1) Calculate the activation energy in kJ/mol for the following reaction if the rate constant for the reaction increases from 93.5 M-1s-1 at 497.7 K to 1349.3 M-1s-1 at 636.7 K. do not include units, but make sure your answer is in kJ/mol! 2) A chemist constructs a plot of ln k vs. 1/T for a chemical reaction. The slope of the trendline for the data is -746 K. What is the activation energy for this reaction in kJ/mol? R...
A chemical reaction was conducted at several different temperatures (measured in K) and the equilibrium constant was measured at each one. Plotting the data on a graph with ln K on the y-axis and 1/T on the x-axis gives a linear plot with slope = -4594.6 and y-intercept = 13.6. What is ΔrH° (in kJ mol-1) and ΔrS° (in J mol-1 K-1)?
The rate constant (k) for a reaction was measured as a function of temperature. A plot of lnk versus 1/T(in K) is linear and has a slope of −1.12×104 K . Calculate the activation energy for the reaction. Express your answer using three significant figures with the appropriate units.
The rate constant (k) for a reaction was measured as a function of temperature. A plot of lnk versus 1/T(in K) is linear and has a slope of -7893 K . Part A Calculate the activation energy for the reaction. Express your answer to four significant figures and include the appropriate units.
The rate constant (k) for a reaction was measured as a function of temperature. A plot of lnk versus 1/T(in K) is linear and has a slope of −1.30×104 K . You may want to reference (Page) Section 14.5 while completing this problem. Calculate the activation energy for the reaction. Express your answer using three significant figures with the appropriate units.