Question

Combustion of octane takes place in an automobile engine. The homogenous equation of combustion is:
${{C}_{8}}{{H}_{18}}(g)+\dfrac{25}{2}{{O}_{2}}(g)\to 8C{{O}_{2}}(g)+9{{H}_{2}}O(g)$
The signs of $\Delta$H, $\Delta$S and $\Delta$G for the reaction will be:
[A] +ve, -ve, +ve
[B] -ve, +ve,-ve
[C] –ve, +ve, +ve
[D] +ve, +ve, -ve

Answer 1

For a reaction to be spontaneous, the change in free energy should be negative. Change in free energy depends on the enthalpy and entropy of the system.

Complete step by step solution:
In thermodynamics, at constant temperature and pressure the change in free energy of a reaction can be written in terms of change in entropy, temperature and enthalpy. It is given as-
$\Delta G=\Delta H-T\Delta S$
Where $\Delta$G is the free energy,
$\Delta$S is the entropy and T is the temperature, on which entropy depends,
$\Delta$H is the enthalpy.
$\Delta$G can be either positive or negative. But, for a reaction to be spontaneous, the change in free energy should be negative which means there should be release of free energy.
$\Delta$S is the spontaneity or randomness of a reaction. With increase in number of molecules, randomness will increase hence change in entropy increase. It can also be negative or positive.
Positive $\Delta$S means the entropy of product is higher than that of the reactant i.e. the number of molecules in the reactant side is less than that of the product side.
$\Delta$H is the amount of heat released or evolved in a reaction at a constant pressure. A negative value of enthalpy means heat is released in the reaction and positive value of enthalpy means energy was required to carry out the reaction.
From the above discussion we can understand that$\Delta$G has to be negative and for $\Delta$G to be negative $\Delta$S should be positive and $\Delta$H should be negative.
$\Delta G=\Delta H-T\Delta S$
$\Delta G=(-ve)-T(+ve)$
As we can see from the above equation it is clear that if entropy is positive and enthalpy is negative, then there will be a release of free energy during the reaction, which means change in free energy will be negative. Therefore, reaction will be spontaneous.
The reaction given to us is- ${{C}_{8}}{{H}_{18}}(g)+\dfrac{25}{2}{{O}_{2}}(g)\to 8C{{O}_{2}}(g)+9{{H}_{2}}O(g)$
As we know, $\Delta$S is positive here as the total number of molecules on the product side is 17 which is greater than the number of molecules on the reactant side.
As it is a combustion reaction, it means heat was released, which means enthalpy for the reaction is negative i.e. $\Delta$H negative.
As entropy is positive and enthalpy is negative, from the equation given by the law of thermodynamics,
$\Delta G=\Delta H-T\Delta S$
$\begin{aligned} & \Delta G=(-ve)-T(+ve) \\\ & \Rightarrow \Delta G=-ve \\\ \end{aligned}$
Therefore, we get $\Delta$S is positive, $\Delta$H is negative and $\Delta$G is also negative.

Therefore the correct answer is option [B] -ve, +ve,-ve.

Note: It is important here to remember that for a reactant to be spontaneous the change in free energy should be always negative and for that the reaction should be exothermic and the randomness on the product side should be higher. If all of this is followed, the reaction will move in the forward direction.
If $\Delta$G is equal to zero, the reaction is at equilibrium and if $\Delta$G is positive, the reaction moves in the forward direction.