Question

Hydrolysis of sucrose is given by the following reaction.
 \(Sucrose + H_20 \rightleftharpoons Glucose + Fructose\) 
If the equilibrium constant \((K_c)\) is \(2 \times 10^{13}\) at 300K, the value of \(\Delta_r\) \(G^\ominus\) at the same temperature will be :

• A. $-8.314\, Jmol^{-1}K^{-1} \times 300 K \times ln (2\times10^{13})$
• B. $8.314 \,mol^{-1}K^{-1} \times 300 K \times ln (2 \times 10^{13})$
• C. $8.314 \, J \,mol^{-1} 300 K \times ln(3 \times 10^{13})$
• D. $-8.314 \,J\,mol^{-1} K^{-1} \times 300 \,K \times In(4 \times 10^{13})$

Answer 1

The Correct Option is A

To find the change in standard Gibbs free energy, \(\Delta_r G^\ominus\), at equilibrium, we use the relationship between \(\Delta_r G^\ominus\) and the equilibrium constant \(K_c\) given by the equation:

\(\Delta_r G^\ominus = -RT \ln K_c\)

where:

• \(R\) is the universal gas constant, which is \(8.314 \, J \, mol^{-1} \, K^{-1}\).
• \(T\) is the temperature in Kelvin, given as \(300 \, K\).
• \(K_c\) is the equilibrium constant, given as \(2 \times 10^{13}\).

Substituting these values into the equation, we have:

\(\Delta_r G^\ominus = -8.314 \, J \, mol^{-1} \, K^{-1} \times 300 \, K \times \ln(2 \times 10^{13})\)

This equation corresponds to the correct answer option:

$-8.314\, Jmol^{-1}K^{-1} \times 300 K \times ln (2\times10^{13})$

The negative sign in the formula indicates that the reaction is spontaneous under standard conditions since \(\ln(2 \times 10^{13})\) is a positive number.