Ea Calculation from Gradient

[IB`NOT`ez]

I'm having a lot of trouble with calculating activation energy from gradient = -Ea/R

Linked to this post is the graph for temperature & ln k. 

The question is: 

Determine the activation energy, Ea , correct to three significant figures and state its units. [3]

Taking the 2nd and 3rd points (from the very top) for gradient, I calculated (-6.20-(-6.70))/(3.22-3.30) = -Ea/8.314

However in the mark scheme, the gradient is supposed to be what I have calculated, multiplied by a 10^3 (thousand). Where did this thousand come from -- I see there's a 10^-3 on the x-axis, but I'm at a loss as to how I can arrive at the gradient I've calculated but also including times 1 thousand.

Furthermore, I'm also still confused as to on the y-axis, are those k values that have yet to be applied "ln" to (I'm assuming it has been), or are they already lnk values and we don't have to apply ln anymore? On the x-axis, are the numbers already the reciprocal, or do we still need to reciprocal it? Then, how do I apply the divided by 10^-3, followed by K^-1?

Would super appreciate clarification on this.

Thank you for taking the time to read this, and thank you again for any help you can provide.

[kw0573]

Hi. It seems you have some trouble reading graphs. In math you learned about graphs and functions, where usually a graph is y is a function of x. In chemistry, however, we more often see graphs of x and y axes that each are functions of other variables. Here vertical axis is a function of k; horizontal axis is a function of T. We are in practice, plotting a two functions against each other. So yes, each axis has already applied the "ln" or reciprocal function. 

Another characteristics is the unit is labelled on each axis. I won't go into details, but most of the time, ln (variable) is unitless, hence no unit is stated. The 1/T is in one thousands of kelvin inverse. You can safely assume in IB chemistry that "/" precedes the unit, when shown on graphs. For example, at 303.03 kelvin, the reciprocal is 0.0033 Kelvin inverse, which is 3.3 of one thousandth of a kelvin inverse. The x-coordinate plotted is 3.3. When you take the slope, you will be dividing by (thousandths of kelvin inverse) which is equivalent to multiplying by a kilokelvin, which explains the thousand in the mark scheme.

[IB`NOT`ez]

4 minutes ago, kw0573 said:

Hi. It seems you have some trouble reading graphs. In math you learned about graphs and functions, where usually a graph is y is a function of x. In chemistry, however, we more often see graphs of x and y axes that each are functions of other variables. Here vertical axis is a function of k; horizontal axis is a function of T. We are in practice, plotting a two functions against each other. So yes, each axis has already applied the "ln" or reciprocal function. 

Another characteristics is the unit is labelled on each axis. I won't go into details, but most of the time, ln (variable) is unitless, hence no unit is stated. The 1/T is in one thousands of kelvin inverse. You can safely assume in IB chemistry that "/" precedes the unit, when shown on graphs. For example, at 303.03 kelvin, the reciprocal is 0.0033 Kelvin inverse, which is 3.3 of one thousandth of a kelvin inverse. The x-coordinate plotted is 3.3. When you take the slope, you will be dividing by (thousandths of kelvin inverse) which is equivalent to multiplying by a kilokelvin, which explains the thousand in the mark scheme.

Thank you for replying so succinctly. I still have trouble understanding your second paragraph though, would you mind explaining it through a series of equation steps? 

[kw0573]

So let's analyze the point (3.30, -6.68) on the graph. -6.68 is not some direct function of 3.30, as you would expect in math. But rather, we have (f(T), g(k)) = (3.30, -6.68). And we define f(T) = the reciprocal of T, and g(k) = ln (k). The unit of f(T) is in 10^-3K^-1. Or the reciprocal of kilokelvin. Now in the given graph, it's more useful to use (1/T) in light of what is to be calculated, the activation energy. But if you are curious to see what temperature, T, f(T) corresponds to, you just take the reciprocal. (Expand for notes).

Spoiler

The function that takes the output of a function and returns x, is called the "inverse function". For example if F(x) = x + 1, then we define the inverse function G(x) = x - 1, such that G(F(x)) = x. It just so happens that the inverse function for the reciprocal function is itself. But for ln(k), the inverse function is e^k. 

 
1 / (3.30 * 10^-3 K^-1) = 1 / (3.30 * 10^-3) * 1 / K^-1 = 303.03 K. So the x-coordinate of 3.30 (10^-3K^-1) is the function f(T) applied for 303.03K. We don't really need 303.03K in this question, but it just goes to show you how f(T) was applied. 

Then in your calculation when you do,  (-6.20-(-6.70))/(3.22-3.30) you have to consider the units being divided: (unitless) / (10^-3 K^-1) = 10³K = kilokelvin. R (gas contant) is usually is JK^-1mol^-1, not some kK^-1, so multiply out the 10³ in 10³K help to keep the units the same (with activation energy in J/mol). As you become more comfortable with unit analysis. You can use alternative methods you like 

[Msj Chem]

To put it simply, you need to find the gradient of the line which is the change in y over the change in x. Be careful of the units of the axis though. 

The gradient of the line is equal to -Ea/R so multiply the gradient by the universal gas constant to get the activation energy. 

 

Edited August 10, 2016 by Msj Chem