1. Tardigrade
  2. Question
  3. Chemistry
  4. For a reversible reaction: X(g) + 3Y(g) rightharpoons 2Z(g) Δ H = - 40 kJ the standard entropies of X, Y and Z are 60,40 and 50 JK-1 mol-1 respectively. The temperature at which the above reaction attains equilibrium is about :

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Q. For a reversible reaction :
$X_{(g)} + 3Y_{(g)} \leftrightharpoons 2Z_{(g)} $
$\Delta H = - 40\, kJ$ the standard entropies of $X, Y$ and $Z$ are $60,40$ and $50 \,JK^{-1}\, mol^{-1}$ respectively.
The temperature at which the above reaction attains equilibrium is about :

KCETKCET 2006Thermodynamics

Solution:

$X(g)+3 Y(g) \quad 2 Z(g) $

$\Delta S^{\circ}=2 \bar{S}^{\circ}(Z)-\left\{S^{\circ}(X)+3 S^{\circ}(Y)\right\} $

$=2 \times 50-\{60+3 \times 40\}$

$=100-180=-80 J K ^{-1} mol ^{-1} $

Given $ \Delta H^{\circ} =-40 kJ =-40,000\, J $

$\Delta G^{\circ} =\Delta H^{\circ}-T \Delta S^{\circ}$

At equilibrium, $\Delta G^{\circ}=0$

$\therefore \Delta H^{\circ}=T \Delta S^{\circ}$

or $T=\frac{\Delta H^{\circ}}{\Delta S^{\circ}}=\frac{40,000}{80}=500 \,K$