For a certain reaction at 300K, K = 10, then for the same reaction is

Question

For a certain reaction at 300K, K = 10, then $∆ G^{o}$ for the same reaction is _________ $\times 10^{- 1}$ kJ ${mol}^{- 1}$ . (Given R = 8.314 ${JK}^{- 1}$ ${mol}^{- 1}$ ) [2024]

Smart Achievers · Accepted Answer

(57)

$ΔG ° = - 2.303 RT \log_{10} K$

$=-2.303×8.314 J {mol}^{- 1} K^{- 1} \times 300 K \log_{10} 10$

$= - 5744.143 J {mol}^{- 1}$

$= - 5.744 kJ {mol}^{- 1} \approx - 57 \times 10^{- 1} kJ {mol}^{- 1}$

Solution

Q.
For a certain reaction at 300K, K = 10, then $∆ G^{o}$ for the same reaction is _________ $\times 10^{- 1}$ kJ ${mol}^{- 1}$ . (Given R = 8.314 ${JK}^{- 1}$ ${mol}^{- 1}$ ) [2024]

Ans.
(57)

$ΔG ° = - 2.303 RT \log_{10} K$

$=-2.303×8.314 J {mol}^{- 1} K^{- 1} \times 300 K \log_{10} 10$

$= - 5744.143 J {mol}^{- 1}$

$= - 5.744 kJ {mol}^{- 1} \approx - 57 \times 10^{- 1} kJ {mol}^{- 1}$

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Solution

Q. For a certain reaction at 300K, K = 10, then ∆Go for the same reaction is _________ ×10-1 kJ mol-1. (Given R = 8.314 JK-1 mol-1) [2024]

Ans. (57) ΔG°=-2.303 RT log10K =-2.303×8.314 Jmol-1 K-1×300 K log1010 =-5744.143 Jmol-1 =-5.744 kJmol-1≈-57×10-1 kJmol-1

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Q.
For a certain reaction at 300K, K = 10, then $∆ G^{o}$ for the same reaction is _________ $\times 10^{- 1}$ kJ ${mol}^{- 1}$ . (Given R = 8.314 ${JK}^{- 1}$ ${mol}^{- 1}$ ) [2024]

Ans.
(57)

$ΔG ° = - 2.303 RT \log_{10} K$

$=-2.303×8.314 J {mol}^{- 1} K^{- 1} \times 300 K \log_{10} 10$

$= - 5744.143 J {mol}^{- 1}$

$= - 5.744 kJ {mol}^{- 1} \approx - 57 \times 10^{- 1} kJ {mol}^{- 1}$