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3 years ago

Calcutale Grxn for the following equation at 25°C: 4KClO3(s) → 3KClO4(s) KCl(s)

Chemistry

1 answer:

Step2247 [10]3 years ago

5 0

Answer:

-133.2 kJ

Explanation:

Let's consider the following balanced equation.

4 KClO₃(s) → 3 KClO₄(s) + KCl(s)

We can calculate the standard Gibbs free energy of the reaction (ΔG°rxn) using the following expression.

ΔG°rxn = 3 mol × ΔG°f(KClO₄(s)) + 1 mol × ΔG°f(KCl(s)) - 4 mol × ΔG°f(KClO₃(s))

ΔG°rxn = 3 mol × (-303.1 kJ/mol) + 1 mol × (-409.1 kJ/mol) - 4 mol × (-296.3 kJ/mol)

ΔG°rxn = -133.2 kJ

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3 years ago

Suppose that a certain biologically important reaction is quite slow at physiological temperature (37 oC) in the absence of a ca

Oksi-84 [34.3K]

Answer:

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Explanation:

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Therefore,

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3 years ago

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Fynjy0 [20]

Answer:

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