[Aragon]

Hello, quick question here.

I am asked to "Calculate the enthalpy of reaction at 800 K".

I understand that I need to fill in following equation (after integrating the integral), in order to get the enthalpy values of every reactant and product:

[aT + b/2T² - cT^-1 + 2dT^0.5]

My question now would be: Do I simply need to put in 800 for T now to get the right answer?

A friend suggested it might be wrong, as he believes I need to put in (800-298) for T.. what is true?

Thank you so much in advance.

[mjc123]

It depends. What are your integration limits? An integral doesn't have a unique value. The integral of xdx is x²/2 + C, where C is a constant, because d/dx(x²/2 + C) = x, whatever the value of C. If you have a definite integral, such as the integral from 1 to 2 of xdx, the value is [x²/2 + C](x=2) - [x²/2 + C](x=1), and C cancels out.
So if you have had to integrate an expression from T1 to T2, your answer is F(T2) - F(T1). I'm guessing T2 is 800K here, but what was T1?

Actually the answer to your friend's question is no. You need F(T2) - F(T1), not F(T2-T1), which is not the same thing. Evaluate the expression at 800K and at 298K (if that is correct), and subtract.