IIT/JEE Chemistry Practice #28: Gibbs Free Energy

The video tutorial explains how to solve a problem involving Gibbs free energy. It begins by introducing a chemical reaction at 300 Kelvin and provides the necessary data, including standard entropy change and Gibbs free energies of formation for CO2 and CO. The tutorial then demonstrates how to calculate the standard Gibbs free energy change using the products minus reactants method. Finally, it shows how to relate Gibbs free energy, enthalpy, and entropy to solve for the standard change in enthalpy, resulting in a value of -285.4 kJ/mol.

What is the standard entropy change given for the reaction at 300 Kelvin?

• 0.194 kilojoules per mole Kelvin

• -0.194 kilojoules per mole Kelvin

• -0.094 kilojoules per mole Kelvin

• 0.094 kilojoules per mole Kelvin

How do you calculate the standard Gibbs free energy change for a reaction?

• By multiplying the Gibbs free energies of formation of products

• By adding the Gibbs free energies of formation of reactants

• By dividing the Gibbs free energies of formation of products by reactants

• By subtracting the Gibbs free energies of formation of reactants from products

What is the standard Gibbs free energy change for the given reaction?

• -394.4 kilojoules per mole

• -285.4 kilojoules per mole

• -257.2 kilojoules per mole

• -137.2 kilojoules per mole

Which equation relates Gibbs free energy, enthalpy, and entropy?

• Delta G = Delta H - T Delta S

• Delta G = Delta H / T Delta S

• Delta G = Delta H + T Delta S

• Delta G = Delta H * T Delta S

What is the calculated standard change in enthalpy for the reaction?

• -285.4 kilojoules per mole

• -394.4 kilojoules per mole

• -257.2 kilojoules per mole

• -137.2 kilojoules per mole