# Predict the feasibility of a reaction when(i) both delta H and delta S increases.(ii)both delta H and delta S decreases.(iii)delta H increases but delta S decreases.(iv)delta H decreases but delta S increases.

We know,

Where,

A reaction is feasible if $∆G$ is negative.
(i) When both delta H and delta S increases, $∆H$  is positive and $∆S$  is also positive. In this case, for $∆G$ to be negative; T$∆S$ should be more than $∆H$. This is possible when temperature is high. So, the reaction will be feasible at high temperature.
(ii)When both delta H and delta S decreases, $∆H$  is negative and $∆S$  is also negative. In this case, for $∆G$ to be negative; T$∆S$ should be less than $∆H$. This is possible when temperature is low. So, the reaction will be feasible at low temperature.
(iii) When delta H increases but delta S decreases, $∆H$  is positive and $∆S$  is also negative. In this case, $∆G$ cannot be negative any temperature.  So, the reaction will not be feasible at any temperature.
(iv) When delta H decreases but delta S increases, $∆H$  is negative and $∆S$  is positive. In this case,  $∆G$  will be negative at all temperatures.  So, the reaction will be feasible at all temperatures.

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A reaction proceeds to decrease the free energy of the system. (i.e. Gibbs free energy decreases in a spontaneous reaction)

Let me write delta with d

delta G = delta H - T delta S

dG = dH - TdS

the condition of feasibility is dG must be negative, which can be accomplished when enthalpy decreases and entropy increases. Let us look at the following possibilities.

(i) When both enthalpy and entropy increase, dH and TdS will be positive. So in order to make dG = dH - TdS negative, the temperature must be high enough such that dH is smaller than TdS. Hence in this case, reactions occuring at high temperatures will be feasible.

(ii) When both H and S decrease, dH will be negative but -TdS will be positive. So the reaction will be feasible at low temperatures.

(iii) When enthalpy increases and entropy decrease, dH - TdS will be positive for all temperatures and hence reaction cannot be feasible.

(iv) In this case, dH will be negative and dS will be positive, such that dH - TdS will be negative for all temperatures and the reaction will be feasible.

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