- #1

Coastal

- 14

- 0

## Homework Statement

For the system 2 CO2 = 2 CO + O2, ∆H= 510 kJ

the percentage decomposition of CO2 changes w/ temperature as follows.

Temp, K...% Decomposition

1500...0.048

2500...17.6

3000...54.8

Calculate the equilibrium constants, plot lnK vs. 1/T. In the graph, find the slope and confirm ∆H. Also, using average bond enthalpies, calculate ∆H(it should come out the same for both approaches).

## Homework Equations

1.) Slope = ∆H / R

2.) R(constant) = 8.314 J/k-mol

3.) ∆Hrxn = Sum of bond enthalpies of broken bonds - Sum of bond enthalpies of formed bonds

## The Attempt at a Solution

I started by calculating K at each temperature. We also did this part in class so I know all the K's are correct.

K at 1500=5.5 x 10-9

K at 2500=4.01 x 10-1

K at 3000=4.03 x 10

I plugged all these #s into the calculator to figure out lnK and 1/T. I plotted them on a graph and got a straight line with a negative slope. This threw me because lnK vs. 1/T should not be straight because this isn't a first order reaction. The slope was something like -40,000, which would then be multiplied by the constant R and ∆H came out to be -330,000(wrong).

Then I tried the 2nd approach. Using equation #3 above, I got ∆H= (4x799)-((2x1072)+495). I'm well aware of how to draw Lewis structures, and am positive I got all the bond enthalpies right and ∆H came out to be 557kJ, which is much closer to the 510kJ given in the problem description.

Where I am stuck is getting ∆H through the graph. Supposedly ∆H=Slope x R(constant)... but with such a small different in 1/T's or the ∆x value or the denominator in the rise/run slope formula, the slope always comes out enormous. How am I supposed to get ∆H for the first method? I looked over the numbers many times and they all seem correct.

Any help please?