Question

The free energy for a reaction having $\Delta H = 31400cal$ , $\Delta S = 32cal$ ${K^{ - 1}}mo{l^{ - 1}}$ at ${1000^ \circ }C$ is:
(A) $ - 9336 \ cal$
(B) $ - 7386 \ cal$
(C) $ - 1936 \ cal$
(D) $ + 9336 \ cal$

Answer 1

Hint: Here in this question, we have to find the free energy of the reaction for which we have to calculate the resultant for the change in enthalpy as with the collision of times the change in entropy as by which we get the resultant as the free energy present in the reaction.

Complete Step by Step Answer:
Before starting the solution we write the given value in the question,
$\Delta H = 31400 \ cal$
$\Delta S = 32 \ cal$

At the temperature of ${1000^ \circ }C$ .
By changing temperature of celsius to kelvin,
Temperature (in kelvin) = $(1000 + 273)k = 1273k$

Now we know that, to find free energy, we use the following formula,
$\Delta G = \Delta H - T\Delta S$
By putting all the values in above equation, we get,
$\Delta G = 31400 - (1273 \times 32)$
By doing further solution we get,
$\Delta G = - 9336 \ cal$
Therefore, the free energy for the reaction is $\Delta G = - 9336 \ cal$ .
Hence, the correct answer is (A).

Note: The total of a chemical reaction's enthalpy (H) and the sum of the system's temperature and entropy (S) is the amount of energy that can be employed to perform work. The kilojoule is the SI unit for Gibbs energy. For processes operating at constant temperature and pressure, variations in the Gibbs free energy G are reflected in variations in the free energy.