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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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Water gas is fuel made by passing steam over red hot coke as $ H_2O(g) + C(s) \rightarrow CO(g)+H_2(g)$ \[\] Given that $ \Delta H_f^{\circ} \: of\: H_2O(g), CO(g)\: are\: -241.8\: and \: -110.5\: kJ/mol$. Also, the values of $S^{\circ}$ for $H_2O(g), C(s), CO(g)\: and\: H_2(g)\: are\: 188.7, 197.9, 131\: and\: 5.69\: J/Kmol$ respectively. Assuming $ \Delta H^{\circ}\: and\: \Delta S^{\circ}$ to be temperature independent, the temperature at which the reaction becomes favorable is

$\begin {array} {1 1} (A)\;994 \: K & \quad (B)\;-994 \: K \\ (C)\;976 \: K & \quad (D)\;967 \: K \end {array}$

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$ \Delta H = [-110.5 + 0] – [-241.8 + 0] = 131.3\: kJ$
$ \Delta S = [197.9 + 131] – [189.7 + 5.69] = 134.55\: J/K$
At equilibrium, $ \Delta H = T \Delta S$
$ \Rightarrow T = \large\frac{\Delta H}{ \Delta S}$$ = 1313 \times \large\frac{103}{134.55}$$ = 967 K$
Thus, the reaction would be spontaneous above $967\: K$
Ans : (D)
answered Mar 19, 2014 by thanvigandhi_1
 

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