One important industrial process to obtain hydrogen gas is by steam reforming of
ID: 917772 • Letter: O
Question
One important industrial process to obtain hydrogen gas is by steam reforming of natural gas at temperatures 700-1100 degree C. An important chemical equilibrium in this process is the equilibrium between methane and water vapor with carbon monoxide and hydrogen gas. Write the chemical equilibrium between methane and water vapor as reactants and carbon monoxide and hydrogen gas as products, using integer coefficients. Evaluate Delta H degree (kJ), DeltaS degree (J/K) at 298.15 K for the equilibrium in part (a). Assuming that Delta H degree and Delta S degree do not change significantly with temperature, for what temperatures (in degree C) is the equilibrium in part (a) spontaneous from left-to-right? Using the values of Delta H degree and Delta S degree from part (a), estimate the values of the equilibrium constant for the equilibrium in part (a) at 400 degree C and 900 degree C. How does your result in part (d) demonstrate LeChatelier's principle?Explanation / Answer
(a) Chemical equation : CH4(g) + H2O(g) <===> CO(g) + 3H2(g)
(b) dHorxn = dHoproducts - dHoreactants
= (-110.525) - (-74.81 - 241.818)
= 206.103 kJ/mol
dSorxn = dSoproducts - dSoreactants
= (197.674 + 130.684) - (186.264 + 188.825)
= -46.731 J/K.mol
(c) T = dHorxn/dSorxn
= 206.103/(-0.04673)
= -4137.51 oC
So below -4137.51 oC the reaction would be spontaneous from left-to-right
(d) dHorxn - TdSorxn = -RTlnKc
T = 400 oC = 400 + 273 = 673 K
206.103 - (400+273)(-0.04673) = -8.314 x (400 x 273) lnKc
Kc = 0.96
T = 900 oC = 900 + 273 = 1173 K
206.103 - (1173)(-0.04673) = -8.314 x 1173 lnKc
Kc = 0.97
(e) At higher temperature the rate of reaction is higher than the reaction at low temperature since enthalpy of reaction is +ve. the reaction is endothermic and more heat added favors the reaction and product formation.
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