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Answers
Explanation:
Predicting the Feasibility of a Possible Redox reaction. with Eo=+0.34V. The more negative the E° value, the further the position of equilibrium lies to the left. ... The equilibrium with the more negative (or less positive) E° value will move to the left.
Answer:
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Explanation:
We know,
∆G = ∆H − T∆S
Where,
∆G is the change in Gibbs free enrgy∆H is tehy change in enthalpy∆S is the change in entropy
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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