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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Thermodynamics

Standard entropy of $X_2, Y_2\: and\: XY_3$ are 60, 40 and 50 J/Kmol respectively. For the reaction, \[\] $ \large\frac{1}{2}$$ X_2 + Y_2 \rightarrow XY_3;\: \: \: \: \: \: \: \: \Delta H = -30\: kJ,$ \[\] to be at the equilibrium, the temperature will be

$\begin {array} {1 1} (A)\;1250\: K & \quad (B)\;500\: K \\ (C)\;750\: K & \quad (D)\;1000\: K \end {array}$

1 Answer

$ \Delta S_{reaction} = s_{products} - S_{reactants}$
$= 50 –(40 + \large\frac{1}{2}$$ \times 60)= -40\: J/mol$
$ \Delta G = \Delta H – T \Delta S$
At equilibrium, $ \Delta G = 0,$
$ \Delta H – T \Delta S$
$T = \large\frac{ \Delta H}{ \Delta S}$$ = \large\frac{(30 \times103 )}{40}$$ = 750\: K$
Ans : (C)
answered Mar 18, 2014 by thanvigandhi_1
 

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