Question

question 3 what is the standard free energy change for a reaction where δh°rxn = -201×10³ j and δs°rxn = -180.0 j/k? 147.3 kj -147.3 kj 266.4 kj -266.4 kj question 4 what is the standard free energy of formation for any pure element under standard conditions? 0 kj 100 kj -100 kj 1 kj question 5 what is the net free energy change for a 3 - step reaction having δg°1 = 111.8 kj, δg°2 = 55.8 kj, δg°3 = -198.8 kj?

Explanation:

Question 3:

Step1: Recall the formula for standard - free energy change

The formula for the standard - free energy change is $\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}$. At standard conditions, $T = 298\ K$. First, convert $\Delta H^{\circ}=-201\times10^{3}\ J$ to $- 201\ kJ$.

Step2: Substitute the values into the formula

$\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}$. Substitute $\Delta H^{\circ}=-201\ kJ$, $T = 298\ K$, and $\Delta S^{\circ}=-180.0\ J/K=-0.1800\ kJ/K$ into the formula.
$\Delta G^{\circ}=-201\ kJ-(298\ K\times(-0.1800\ kJ/K))$.

Step3: Calculate the result

First, calculate $298\times(-0.1800)=-53.64$. Then, $\Delta G^{\circ}=-201\ kJ + 53.64\ kJ=-147.36\ kJ\approx - 147.3\ kJ$.

Question 4:

The standard free - energy of formation of any pure element under standard conditions is $0\ kJ$.

Question 5:

Step1: Sum up the free - energy changes of each step

For a multi - step reaction, the net free - energy change $\Delta G_{net}=\Delta G_{1}+\Delta G_{2}+\Delta G_{3}$. Given $\Delta G_{1}=111.8\ kJ$, $\Delta G_{2}=55.8\ kJ$, and $\Delta G_{3}=-198.8\ kJ$.

Step2: Calculate the net free - energy change

$\Delta G_{net}=111.8\ kJ + 55.8\ kJ-198.8\ kJ=(111.8 + 55.8-198.8)\ kJ=-31.2\ kJ$.

Answer:

Question 3: $-147.3\ kJ$
Question 4: $0\ kJ$
Question 5: $-31.2\ kJ$