For the reaction 2NOg- O 2 g- 2N O 2 gCalculate - G at 700 K when enthalpy and entropy changes are -113 kJ mol -1 and -145 J K -1 mol -1 respectively-

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Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

For the react...

Question

For the reaction
$2 N O (g) + O_{2} (g) ⟶ 2 N O_{2} (g)$
Calculate $Δ G$ at $700 K$ when enthalpy and entropy changes are $- 113 k J {m o l}^{- 1}$ and $- 145 J K^{- 1} {m o l}^{- 1}$ respectively.

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Solution

We know that, $Δ G = Δ H - T Δ S$
Given that, $Δ H = - 113 k J {m o l}^{- 1} = - 113000 J {m o l}^{- 1}$
$Δ S = - 145 J K^{- 1} {m o l}^{- 1}$
$T = 700 K$
Substituting these values in the above equation
$G = - 113000 - 700 \times (- 145) = - 11500 J {m o l}^{- 1} = - 11.5 k J {m o l}^{- 1}$

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For the reaction2NO(g)+O2(g)⟶2NO2(g)Calculate ΔG at 700 K when enthalpy and entropy changes are −113 kJmol−1 and −145 JK−1mol−1 respectively.

We know that, ΔG=ΔH−TΔSGiven that, ΔH=−113kJmol−1=−113000Jmol−1ΔS=−145JK−1mol−1T=700KSubstituting these values in the above equationG=−113000−700×(−145)=−11500Jmol−1=−11.5kJmol−1

For the reaction
$2 N O (g) + O_{2} (g) ⟶ 2 N O_{2} (g)$
Calculate $Δ G$ at $700 K$ when enthalpy and entropy changes are $- 113 k J {m o l}^{- 1}$ and $- 145 J K^{- 1} {m o l}^{- 1}$ respectively.

We know that, $Δ G = Δ H - T Δ S$
Given that, $Δ H = - 113 k J {m o l}^{- 1} = - 113000 J {m o l}^{- 1}$
$Δ S = - 145 J K^{- 1} {m o l}^{- 1}$
$T = 700 K$
Substituting these values in the above equation
$G = - 113000 - 700 \times (- 145) = - 11500 J {m o l}^{- 1} = - 11.5 k J {m o l}^{- 1}$