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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}$$X_2 + Y_2 \rightarrow XY_3;\: \: \: \: \: \: \: \: \Delta H = -30\: kJ,  to be at the equilibrium, the temperature will be \begin {array} {1 1} (A)\;1250\: K & \quad (B)\;500\: K \\ (C)\;750\: K & \quad (D)\;1000\: K \end {array} Can you answer this question? ## 1 Answer 0 votes \Delta S_{reaction} = s_{products} - S_{reactants} = 50 –(40 + \large\frac{1}{2}$$ \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)