Question

What will be the entropy change in surrounding when 1 mol of ${{H}_{2}}O$is formed under standard conditions: $\Delta H{{{}^\circ }_{f}}=-286KJ/mol$?
A. 400 J/K
B. 0
C. 959.7 K
D. 37 J/K

Answer 1

Entropy is a thermodynamic function. Whereas $\Delta S$ defines the change in free energy state which is given by the change in the enthalpy of the system minus the product of temperature and change in the entropy which is given by:
$\Delta G=\Delta H-T\Delta S$

Complete answer:
Entropy is defined as a randomness or disorder of the system as the particles adsorb the randomness of the particles goes on decreasing and the process becomes nonspontaneous in nature and non-spontaneous process is negative in nature.
In the given question value of $\Delta H{{{}^\circ }_{f}}=-286KJ/mol$is given where this implies the heat evolved in formation of one mole of water. This will be absorbed by the surroundings. So this will implies that heat absorbed will also be equal to the value of heat formation given i.e.
${{Q}_{surrounding}}=+286KJmo{{l}^{-1}}$
Then the change in entropy will be calculated by the following formula
$\Delta {{S}_{surrounding}}=\dfrac{{{Q}_{surrounding}}}{T}$
Here T stands for the temperature, temperature is not given but standard conditions are mentioned so according to standard conditions the temperature will be 298 K, now put the values of Q and T in the formula:
$\Delta {{S}_{surrounding}}=\dfrac{286}{298}=959.73J/K$

This suggests that option C is the correct answer.

Note:
If H i.e. enthalpy is negative and S i.e. entropy is positive then the sign of G will also be negative and at both temperatures the reaction will be random in nature. This concludes that all driving forces are in favour of product formation.