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(b) Given that ΔG° for reaction 1 is positive (155kJ/molrxn), what must be true about the sign of ΔH° for the reaction? Justify your answer

Sagot :

Answer: The sign of ΔH° for the reaction is positive.

Explanation:

Given: [tex]\Delta G^{o}[/tex] = 155 kJ/mol (1 kJ = 1000 J) = 155000 J/mol

The relation between [tex]\Delta G^{o}[/tex] and [tex]\Delta H^{o}[/tex] is as follows.

[tex]\Delta G^{o} = \Delta H^{o} - T \Delta S[/tex]

As the given value of [tex]\Delta G^{o}[/tex] is positive. This means that on further calculation the value of [tex]\Delta H^{o}[/tex] will come out to be positive.

Thus, we can conclude that the sign of ΔH° for the reaction is positive.

In a closed system, the amount of the non-expansion work that can be done is called gibs free energy. The enthalpy cahnge will be positive.

What is enthalpy?

The enthalpy of the system is given as the addition of products of the pressure and the volume with the internal energy of the system.

The Gibbs free energy is given by the formula:

[tex]\rm \Delta G^{\circ} = \Delta H ^{\circ} - T\Delta S[/tex]

When Gibbs's free energy is positive then the enthalpy will become positive.

Therefore, the enthalpy change will be positive.

Learn more about Gibbs free energy here:

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