1. Tardigrade
  2. Question
  3. Chemistry
  4. In the reaction: A2(g) + 3B2(g) → 2AB3(g) the standard entropies (in JK-1 mol-1) of A2(g), B2(g) and AB3(g) are respectively 190, 130 and 195 and the standard enthalpy change for the reaction is -95 kJ mol-1. The temperature (in K) at which the reaction attains equilibrium is (assuming both the standard entropy change and standard enthalpy change for this reaction are constant over a wide range of temperature)

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Q. In the reaction: $A_{2\left(g\right)} + 3B_{2\left(g\right)} \to 2AB_{3\left(g\right)}$
the standard entropies (in $JK^{-1}$ $mol^{-1}$) of $A_{2(g)}$, $B_{2(g)}$ and $AB_{3(g)}$ are respectively $190, 130$ and $195$ and the standard enthalpy change for the reaction is $-95\, kJ \,mol^{-1}$. The temperature ($in\, K$) at which the reaction attains equilibrium is (assuming both the standard entropy change and standard enthalpy change for this reaction are constant over a wide range of temperature)

KEAMKEAM 2012Thermodynamics

Solution:

For the reaction,

$A_{2}(g)+3 B_{2}(g) \longrightarrow 2 A B_{3}(g)$

Entropy change, $\Delta S^{\circ}=\Sigma S_{\text {product }}-\Sigma S_{\text {reactants }}$

$=2 \times S_{A B_{3}}-\left[S_{A_{2}}+3 S_{B_{2}}\right]$

$=2 \times 195-[190+3 \times 130] $

$=390-[190+390] $

$=-190 JK ^{-1} mol ^{-1}$

Given, $\Delta H^{\circ}=-95\, kJ\, mol ^{-1}$

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

We know that,

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

$0 =\Delta H^{\circ}-T \Delta S^{\circ} $

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

$T=\frac{\Delta H^{\circ}}{\Delta S^{\circ}} =\frac{-95 \times 10^{3} J mol ^{-1}}{-190 J K ^{-1} mol ^{-1}}$ $=500 K$