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What is the value of [tex]\Delta G[/tex] at [tex]300\, K[/tex] if [tex]\Delta H = 27\, kJ/mol[/tex] and [tex]\Delta S = 0.09\, kJ/(mol \cdot K)[/tex]?

A. [tex]\Delta G = 27\, kJ/mol[/tex]

B. [tex]\Delta G = 54\, kJ/mol[/tex]

C. [tex]\Delta G = 0\, kJ/mol[/tex]

D. [tex]\Delta G = -18\, kJ/mol[/tex]

Sagot :

GinnyAnsweridn
To solve for the Gibbs free energy change ([tex]\(\Delta G\)[/tex]), we can use the Gibbs free energy equation:

[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]

where:
- [tex]\(\Delta H\)[/tex] is the change in enthalpy,
- [tex]\(T\)[/tex] is the temperature in Kelvin,
- [tex]\(\Delta S\)[/tex] is the change in entropy.

Given:
- [tex]\(\Delta H = 27 \, \text{kJ/mol}\)[/tex],
- [tex]\(\Delta S = 0.09 \, \text{kJ/(mol \cdot K)}\)[/tex],
- [tex]\(T = 300 \, \text{K}\)[/tex].

First, we substitute these values into the equation:

[tex]\[ \Delta G = 27 \, \text{kJ/mol} - 300 \, \text{K} \times 0.09 \, \text{kJ/(mol \cdot K)} \][/tex]

Next, perform the multiplication:

[tex]\[ 300 \, \text{K} \times 0.09 \, \text{kJ/(mol \cdot K)} = 27 \, \text{kJ/mol} \][/tex]

Then, subtract this result from the enthalpy change:

[tex]\[ \Delta G = 27 \, \text{kJ/mol} - 27 \, \text{kJ/mol} \][/tex]

[tex]\[ \Delta G = 0 \, \text{kJ/mol} \][/tex]

Thus, the value for [tex]\(\Delta G\)[/tex] at [tex]\(300 \, \text{K}\)[/tex] is:

C. [tex]\(\Delta G = 0 \, \text{kJ/mol}\)[/tex]
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