Use the Gibbs function to determine the equilibrium constant of the H 2 O H 2
Question:
Use the Gibbs function to determine the equilibrium constant of the H2O ⇌ H2 + 1/2 O2 reaction at
(a) 1440 R
(b) 3960 R.
How do these compare to the equilibrium constants of Table A–28?
Transcribed Image Text:
TABLE A-28 Natural logarithms of the equilibrium constant K, The equilibrium constant K, for the reaction vA + v„B = vC + vpD is defined as K, = PP Temp., K H, = 2H 0, 20 N2 = 2N H,0 = H, + ½0, H,O = ½H, + OH CO, CO + ½0, ½N, + ½0, NO 298 500 1000 -164.005 -186.975 -367.480 -92.208 -106.208 - 103.762 -35.052 -92.827 -105.630 -45.150 -35.005 -27.742 -213.372 -52.691 -60.281 -57.616 -20.295 -39.803 -30.874 -99.127 -80.011 -9.388 -7.569 -23.163 -26.034 -20.283 -23.529 -17.871 -13.842 1200 -18.182 -6.270 -5.294 1400 -24.463 -66.329 -14.609 1600 1800 -19.637 -15.866 -22.285 -18.030 -14.622 - 16.099 -13.066 - 10.657 -56.055 -48.051 -11.921 -9.826 - 10.830 -8.497 -6.635 -5.120 -4.536 2000 2200 -12.840 - 10.353 -41.645 -8.145 -6.768 -8.728 -7.148 -3.931 -3.433 -11.827 -36.391 2400 -8.276 -9.497 -32.011 -5.619 -4.648 -3.812 -5.832 -3.860 -3.019 -28.304 2600 2800 -6.517 -5.002 -7.521 -5.826 -4.719 -3.763 -2.801 -1.894 -2.671 -2.372 -25.117 -22.359 3000 3200 -3.685 -2.534 -4.357 -3.072 -3.086 -2.937 -2.212 -1.111 -0.429 -2.114 -1.888 -19.937 -2.451 3400 -1.516 -1.935 -17.800 -1.891 -1.576 0.169 3600 3800 -0.609 0.202 -0.926 -0.019 -15.898 -14.199 -12.660 -1.690 -1.513 -1.356 -1.392 -0.945 -1.088 -0.501 0.701 1.176 0.796 2.513 3.895 0.934 -0.542 -0.044 0.920 1.689 4000 1.599 -1.216 4500 5000 2.486 3.725 -9.414 -6.807 0.312 2.490 3.197 -0.921 -0.686 0.996 5500 4.743 5.023 -4.666 1.560 2.318 3.771 -0.497 6000 5.590 5.963 -2.865 2.032 2.843 4.245 -0.341
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Related Book For
Thermodynamics An Engineering Approach
ISBN: 9781259822674
9th Edition
Authors: Yunus Cengel, Michael Boles, Mehmet Kanoglu
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