Answer :
To find the value of [tex]\(\Delta G\)[/tex] given that [tex]\(\Delta H = 27 \text{ kJ/mol}\)[/tex], [tex]\(\Delta S = 0.09 \text{ kJ/(mol·K)}\)[/tex], and the temperature [tex]\(T = 300 \text{ K}\)[/tex], we can use the Gibbs free energy equation:
[tex]\[\Delta G = \Delta H - T\Delta S\][/tex]
Now, let's substitute the given values into this equation:
[tex]\[ \Delta G = 27 \text{ kJ/mol} - (300 \text{ K} \times 0.09 \text{ kJ/(mol·K)}) \][/tex]
First, calculate the product [tex]\(T\Delta S\)[/tex]:
[tex]\[ T\Delta S = 300 \text{ K} \times 0.09 \text{ kJ/(mol·K)} = 27 \text{ kJ/mol} \][/tex]
Now, substitute [tex]\(T \Delta S\)[/tex] back into the equation for [tex]\(\Delta G\)[/tex]:
[tex]\[ \Delta G = 27 \text{ kJ/mol} - 27 \text{ kJ/mol} = 0 \text{ kJ/mol} \][/tex]
Therefore, the value for [tex]\(\Delta G\)[/tex] is:
[tex]\[ \Delta G = 0 \text{ kJ/mol} \][/tex]
The correct answer is:
B. [tex]\(\Delta G = 0 \text{ kJ/mol}\)[/tex]
[tex]\[\Delta G = \Delta H - T\Delta S\][/tex]
Now, let's substitute the given values into this equation:
[tex]\[ \Delta G = 27 \text{ kJ/mol} - (300 \text{ K} \times 0.09 \text{ kJ/(mol·K)}) \][/tex]
First, calculate the product [tex]\(T\Delta S\)[/tex]:
[tex]\[ T\Delta S = 300 \text{ K} \times 0.09 \text{ kJ/(mol·K)} = 27 \text{ kJ/mol} \][/tex]
Now, substitute [tex]\(T \Delta S\)[/tex] back into the equation for [tex]\(\Delta G\)[/tex]:
[tex]\[ \Delta G = 27 \text{ kJ/mol} - 27 \text{ kJ/mol} = 0 \text{ kJ/mol} \][/tex]
Therefore, the value for [tex]\(\Delta G\)[/tex] is:
[tex]\[ \Delta G = 0 \text{ kJ/mol} \][/tex]
The correct answer is:
B. [tex]\(\Delta G = 0 \text{ kJ/mol}\)[/tex]
