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It was measured that 25 % of the total volume of 3 kmol of diatomic hydrogen gas (H2) is dissociated into monatomic hydrogen (H). Consider
It was measured that 25 % of the total volume of 3 kmol of diatomic hydrogen gas (H2) is dissociated into monatomic hydrogen (H). Consider the reaction pressure to be 5.1 atm. Use the attached table below to determine the temperature. (i) Write the balanced molar reaction of dissociation. (ii) Calculate the temperature at which the reaction equilibrium occurs.
Natural logarithms of the equilibrium constant K, PEP The equilibrium constant K, for the reaction v1A + vgB = vcC + voD is defined as K, PX PX Temp., H2 = 24 O2 = 20 N2 = 2N H20 =H2 + 1,0 H2O="/2H2 + OH CO=CO+ "/202N2 + 1/2O2 = NO 298 -164.005 -186.975 -367.480 -92.208 -106.208 - 103.762 -35.052 500 -92.827 -105.630 -213.372 -52.691 -60.281 -57.616 -20.295 1000 -39.803 -45.150 -99.127 -23.163 -26.034 -23.529 -9.388 1200 -30.874 -35.005 -80.011 -18.182 -20.283 -17.871 -7.569 1400 -24.463 -27.742 -66.329 - 14.609 - 16.099 -13.842 -6.270 1600 - 19.637 -22.285 -56.055 -11.921 -13.066 -10.830 -5.294 1800 - 15.866 -18.030 -48.051 -9.826 -10.657 -8.497 -4.536 2000 -12.840 -14.622 -41.645 -8.145 -8.728 -6.635 -3.931 2200 -10.353 -11.827 -36.391 -6.768 -7.148 -5.120 -3.433 2400 -8.276 -9.497 -32.011 -5.619 -5.832 -3.860 -3.019 2600 -6.517 -7.521 -28.304 -4.648 -4.719 -2.801 -2.671 2800 -5.002 -5.826 -25.117 -3.812 -3.763 -1.894 -2.372 3000 -3.685 -4.357 -22.359 -3.086 -2.937 -1.111 -2.114 3200 -2.534 -3.072 -19.937 -2.451 -2.212 -0.429 -1.888 3400 -1.516 -1.935 -17.800 -1.891 -1.576 0.169 -1.690 3600 -0.609 -0.926 - 15.898 -1.392 -1.088 0.701 -1.513 3800 0.202 -0.019 - 14.199 -0.945 -0.501 1.176 -1.356 4000 0.934 0.796 -12.660 -0.542 -0.044 1.599 -1.216 4500 2.486 2.513 -9.414 0.312 0.920 2.490 -0.921 5000 3.725 3.895 -6.807 0.996 1.689 3.197 -0.686 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 -0.341Step by Step Solution
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