Assume H 0 and S 0 to be independent of temperature- at what temperature will the reaction given below becomes spontaneous- N 2 g - O 2 g - 2 NO g S N 2 g 0-191-4 JK -1 mol -1S O 2 g 0-204-9 JK -1 mol -1S NO g 0-210-5 JK -1 mol -1H -180-8 kJ mol -1A- 7320 KB- 3650 KC- 6595 KD- 8500 K

The correct option is A 7320 KGiven, nbsp;N_2(g) + O_2(g)→2NO(g) nbsp; S^o_N_2(g)=191.4 JK^ 1 mol^ 1 S^o_O_2(g)=204.9 JK^ 1 mol^ 1 S^o_NO(g)=210.5 JK^ 1 mo ...

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Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

Assume H 0 an...

Question

Assume $△ H^{o} a n d △ S^{o}$ to be independent of temperature, at what temperature will the reaction given below becomes spontaneous?
$N_{2} (g) + O_{2} (g) \to 2 N O (g)$
$S_{N_{2} (g)}^{o} = 191.4 J K^{- 1} m o l^{- 1}$
$S_{O_{2} (g)}^{o} = 204.9 J K^{- 1} m o l^{- 1}$
$S_{N O (g)}^{o} = 210.5 J K^{- 1} m o l^{- 1}$
$△ H^{o} = 180.8 k J m o l^{- 1}$

7320 K

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3650 K

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6595 K

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8500 K

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Solution

The correct option is A 7320 K
Given, $N_{2} (g) + O_{2} (g) \to 2 N O (g)$
$S_{N_{2} (g)}^{o} = 191.4 J K^{- 1} m o l^{- 1}$
$S_{O_{2} (g)}^{o} = 204.9 J K^{- 1} m o l^{- 1}$
$S_{N O (g)}^{o} = 210.5 J K^{- 1} m o l^{- 1}$
$△ H^{o} = 180.8 k J m o l^{- 1}$
The entropy of the reaction can be calculated as
$△ S^{o} = (2 S_{N O}^{o}) - (S_{N_{2}}^{o} + S_{O_{2}}^{o})$
putting the values, we get
$= (2 \times 210.5) - (191.4 + 204.9)$
$= 24.7 J K^{- 1} m o l^{- 1}$
Also, we know gibb's free energy relation,
$△ G^{o} = △ H^{o} - T △ S^{o}$
Putting the values, we get;
$△ G^{o} = 180.8 - (T \times 24.7 \times 10^{- 3})$
Reaction becomes spontaneous, when
$△ G^{o} < 0$
$180.8 - T \times 24.7 \times 10^{- 3} < 0$
which means,
$T > \frac{180.8}{24.7} \times 10^{3} = 7320 K$

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Similar questions

Q. Assume

△ H^{o} a n d △ S^{o}

to be independent of temperature, at what temperature will the reaction given below becomes spontaneous?

N_{2} (g) + O_{2} (g) \to 2 N O (g)

S_{N_{2} (g)}^{o} = 191.4 J K^{- 1} m o l^{- 1}

S_{O_{2} (g)}^{o} = 204.9 J K^{- 1} m o l^{- 1}

S_{N O (g)}^{o} = 210.5 J K^{- 1} m o l^{- 1}

△ H^{o} = 180.8 k J m o l^{- 1}

Q. For a chemical reaction

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(Δ_{r} H^{0} = 80 k J m o l^{- 1})

the entropy change

Δ_{r} S^{0}

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Δ_{r} S^{0} = 2 T (J K^{- 1} m o l^{- 1})

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For the reaction at 298 K,

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Gibbs Free Energy & Spontaneity

Standard XII Chemistry

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Assume △ Ho and △ So to be independent of temperature, at what temperature will the reaction given below becomes spontaneous? N2(g)+O2(g)→2NO(g) SoN2(g)=191.4 JK−1 mol−1 SoO2(g)=204.9 JK−1 mol−1 SoNO(g)=210.5 JK−1 mol−1 △Ho=180.8 kJ mol−1