In: Chemistry
You proceed to measure the equilibrium constant at three more temperatures, and now have the following results:
Temperature (K) |
Equilibrium constant |
100.0 ± 0.01 |
(1.6±0.1) ×104 |
150.0±0.01 |
(3.3±0.1) ×104 |
200.0 ± 0.01 |
(5.7±0.1) ×104 |
250.0±0.01 |
(6.2±0.1) ×104 |
300.0±0.01 |
(7.8±0.1) ×104 |
(a) Linearize your date and perform a linear regression, using Excel or another computer program. Use the program to find the slope, intercept, and uncertainties in the slope and intercept.
(b) Make a figure of the linearized data in 7(a); be sure to include the best-fit line.
(c) Calculate the ΔH and ΔS for this reaction.
(d) Estimate the uncertainties for ΔH and ΔS in part (c).
(a) Linear data points,
Temperature (K) (1/T)) Equilibrium constant (K) (lnK)
100 (0.01) 1.6x10^4 (9.68)
150 (0.007) 3.3x10^4 (10.40)
200 (0.005) 5.7x10^4 (10.95)
250 (0.004) 6.2x10^4 (11.035)
300 (0.003) 7.8x10^4 (11.26)
To calculate maximum slope we will use the values, (100.01,1.5x10^4) and (299.99,7.9x10^4)
maximum slope = y2-y1/x2-x1 = 7.9x10^4-1.5x10^4/299.99-100.01 = 320.032
To calculate minimum slope values, (99.99,1.6x10^4) and (300.01,7.7x10^4)
Minimum slope = 7.7x10^4-1.6x10^4/300.01-99.99) = 304.9695
uncertainty in the slope = max slope-min slope/2 = 320.032-304.9695/2 = 7.53
So the actual slope of this line becomes = 320 +/- 7.53
The uncertainty in the intercept
Intercept of line is 0.33x10^4
the uncertainty is approximately 0.1
(b) Using the relation,
lnKeq = -deltaH/RT + delta S/R
Plot lnKeq vs 1/T [data given above]
(c) Find delta H and delta S from slope and intercept of the line
slope = -240 = -deltaH/R
deltaH = 240 x 8.314 = 1995.36 J/mol
intercept = deltaS/R = 9.5265
delta S = 9.5265 x 8.314 = 79.203 J/K.mol
(d) Uncertainty for delta H +/-0.01
For delta S = +/-0.01