Question

question 19 of 30

at 298 k, $\delta h^0 = 46$ kj/mol and $\delta s^0 = 0.097$ kj/(k·mol). what is the gibbs free energy of the reaction?

a. 0.63 kj

b. 1300 kj

c. 17 kj

d. 75 kj

Explanation:

Step1: Recall Gibbs free energy formula

The formula for Gibbs free energy change ($\Delta G^0$) is $\Delta G^0=\Delta H^0 - T\Delta S^0$, where $\Delta H^0$ is the enthalpy change, $T$ is the temperature in Kelvin, and $\Delta S^0$ is the entropy change.

Step2: Substitute the given values

We are given $\Delta H^0 = 46\space kJ/mol$, $T = 298\space K$, and $\Delta S^0=0.097\space kJ/(K\cdot mol)$. First, calculate $T\Delta S^0$:
$T\Delta S^0=298\space K\times0.097\space kJ/(K\cdot mol)=298\times0.097\space kJ/mol$.
Calculating $298\times0.097$: $298\times0.097=(300 - 2)\times0.097 = 300\times0.097-2\times0.097=29.1 - 0.194 = 28.906\space kJ/mol$.
Then, calculate $\Delta G^0$:
$\Delta G^0=\Delta H^0 - T\Delta S^0=46\space kJ/mol-28.906\space kJ/mol = 17.094\space kJ/mol\approx17\space kJ/mol$.

Answer:

C. 17 kJ