Question

For a particular reaction, Δ=−111.4 kJ and Δ=−25.0 J/K.

Calculate Δ for this reaction at 298 K.
Δ= ?

Answer 1

`\Delta G =-103.95kJ`

Step-by-step explanation:

Hello there!

In this case, since the thermodynamic definition of the Gibbs free energy for a change process is:

`\Delta G =\Delta H-T\Delta S`

It is possible to plug in the given H, T and S with consistent units, to obtain the correct G as shown below:

`\Delta G =-111.4kJ-(298K)(-25.0(J)/(K)*(1kJ)/(1000J) )\\\\\Delta G =-103.95kJ`

Best regards!