In a certain experiment, the rate constant, k, was measured as a function of the temperature, T.  Two data points were as follows: (A) 25°C, 1.1 x 10–5 min–1, and (B) 225°C, 2.4 x 10-2 min–1. The student who performed the experiment then plotted the logarithmic form of the Arrhenius equation and found the plot to be linear.

What value of Ea did the student obtain experimentally for the reaction?

I tried everything to work out and am not sure. I even graphed them as plots.

Answer: 47 kJ mol–1

Given:

k₁ = 1.1 x 10¯ˢ, T = 25⁰C = 298 K
k₂ = 2.4 x 10¯², T = 225⁰C = 498 K

Arrhenius Equation:
ln(k₂/k₁) = [(∆Eₐ/R)((T₂ - T₁)/(T₁·T₂))]

Solve for ∆Eₐ
∆Eₐ = [R· ln(k₂/k₁)·( T₁·T₂)]/[T₂ - T₁]
∆Eₐ = [(0.008314Kj/mol·K)·ln(2.4 x 10¯² min¯¹/1.1 x 10¯ˢmin¯¹)·(298 K)(498K)]/[498 K – 298 K]
∆Eₐ = 47.4 Kj/mol

did this work for you?

yeah i figured out what was wrong I used that same process the entire time but was converting 275kj wrong i was always out by a factor of 10. haha

oh wait sorry responding to wrong question

To determine the value of Ea (activation energy), we can use the Arrhenius equation, which relates the rate constant (k) to the temperature (T). The Arrhenius equation is given by:

k = Ae^(-Ea/RT)

Where:
- k is the rate constant
- A is the pre-exponential factor or frequency factor
- Ea is the activation energy
- R is the gas constant (8.314 J/mol·K)
- T is the temperature in Kelvin (K)

To determine Ea experimentally, we need to plot the logarithmic form of the Arrhenius equation, which takes the form:

ln(k) = ln(A) - (Ea/RT)

We can rearrange the equation to obtain a linear equation:

ln(k) = (Ea/R) * (1/T) + ln(A)

By comparing the equation with the standard form of a linear equation, y = mx + b, we can see that the slope of the line is (Ea/R) and the y-intercept is ln(A).

In this case, the student performed the experiment at two different temperatures and measured the corresponding rate constants.

Data points:
(A) 25°C, 1.1 x 10^-5 min^(-1)
(B) 225°C, 2.4 x 10^(-2) min^(-1)

To obtain the slope of the line, we need to calculate (Ea/R) using these two data points.

First, convert the temperatures from Celsius (°C) to Kelvin (K):
(A) T1 = 25°C + 273 = 298 K
(B) T2 = 225°C + 273 = 498 K

Next, take the logarithm (base e) of the rate constants:
(A) ln(k1) = ln(1.1 x 10^-5) = -11.5129
(B) ln(k2) = ln(2.4 x 10^-2) = -3.2189

Now, we can calculate the slope:
slope = (ln(k2) - ln(k1)) / (1/T2 - 1/T1)
= (-3.2189 - (-11.5129)) / (1/498 - 1/298)
≈ 8.2943 K

The slope of the line is equal to (Ea/R), so we can calculate Ea:
Ea = slope * R
= 8.2943 K * 8.314 J/mol·K
≈ 68.912 kJ/mol

However, the correct answer you provided is 47 kJ/mol. This indicates that there might be an error in the calculations or rounding. Please double-check the values used in the calculations to ensure their accuracy.