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Equilibrium Constant and Standard Free Energy Change

Calculate fre...

Question

Calculate free energy change for the conversion of oxygen to ozone at 298 K, if $K_{p}$ for the change $\frac{3}{2} O_{2} (g) \to O_{3} (g)$ at 298 K is $2.47 \times 10^{- 29} (a t m)^{- \frac{1}{2}} .$
log 2.47=0.3926

163.23 kJ

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170.14 KJ

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159.24 KJ

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160.00 KJ

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Solution

The correct option is A 163.23 kJ
Given : $T = 298 K, K_{p} = 2.47 \times 10^{- 29} (a t m)^{- \frac{1}{2}}$
We know the relation, $△ G^{o} = - 2.303 R T l o g_{10} K_{p}$
$= - 2.303 \times 8.314 \times 298 \times l o g 2.47 \times 10^{- 29}$
$= 5705.848 \times l o g 2.47 \times 10^{- 29}$
$= 5705.848 \times (- 29 + l o g 2.47$
$= 5705.848 \times (- 29 + 0.3926$
$= 163229 J$
$= 163.229 k J$
$= 163.23 k J$

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Similar questions

Δ G^{\circ}

\frac{3}{2} O_{2_{(g)}} \to O_{3_{(g)}}

K_{p}

2.47 \times 10^{- 29}

\frac{3}{2} O_{2} (g) ⟶ O_{3} (g)

K_{p}

2.47 \times 10^{- 29}

Δ_{r} G^{o}

△ G^{\circ}

3 / 2 O_{2} (g) \to O_{3} (g) a t 298 K, i f K_{p}

\times

10^{- 29}

Δ G^{o}

(3 / 2) O_{2} (g) ⇌ O_{3} (g)

298 K

K_{P}

2.47 \times 10^{- 29}

Δ_{r} G^{o}

O_{2} (g) \to O_{2} (g)

K_{p}

2.47 \times 10^{- 29}

l o g_{10} 2.47 = 0.392

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