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Isn't setting $\Delta G = 0$ akin to enforcing a statement about the spontaneity of the reaction? Yes, it is. But they stipulate that the reaction is at equilibrium, which tells you that neither the forward nor backward reaction is spontaneous, which is precisely what allows you to assert that $\Delta G = 0$.* In other words, when a system is at equilibrium,...


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Thermodynamically if you calculate the equilibrium constant the equilibrium quotient depends on number of moles of both reactants as well the number of moles of products that is reaction concentration at the equilibrium.more the free energy of reactants the reaction will proceed in a spontaneous manner the more feasible the reaction . Likewise the more the ...


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Those are not Coupled reactions; the burning of carbon simply supplies the energy to decompose the carbonate. Another way of looking at it would be if you were the fireman on an old steam locomotive would you prefer to shovel pure coal or a mix of coal and limestone into the firebox for several hours trying to keep the train on time. The reactions are ...


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This is the complete question at the GenChem 2 level (found posted on Chegg): The correct answer is C. Of course, there is a relationship between most concepts, but not one where the speed of a reaction would reliably predict the sign of the standard Gibbs energy. Also, fast reactions is a fuzzy concept. A given reaction will be faster or slower depending ...


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