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29 October, 09:06

A reaction with ΔH = - 13 kJ/mol and ΔS = - 22 J/K·mol, at 2°C, the reaction is:

A) spontaneous

B) nonspontaneous

C) at equilibrium

D) impossible to determine reactivity

E) none of the above

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Answers (1)
  1. 29 October, 10:30
    0
    B

    Explanation:

    The value of the Free Gibbs energy is the indicator used to access of a reaction is spontaneous or not. Hence, what we are saying is that the value of the Gibbs free energy is the deciding factor to note if a reaction is spontaneous or not.

    Theoretically, this value can be accessed using experimental data. By knowing the value in the change in enthalpy, the change in entropy and the temperature condition, we can get the value of the change in Gibbs free energy. The formula for this is written as follows:

    ΔG = ΔH - TΔS

    From the question, we know that change in enthalpy equals - 13KJ/mol. Change in entropy is - 22KJ/mol.

    Temperature is 2 degrees celcius which is same as 2 + 273.15 = 275.15

    Inputting these values into the equation yields the following:

    ΔG = - 13 + 22 (273.15)

    = 5,996

    Since the value is positive, the reaction is not spontaneous.
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