# Question 04ff1

May 15, 2015

The change in free energy will be equal to $\text{+173 kJ}$.

${N}_{2 \left(g\right)} + {O}_{2 \left(g\right)} \to \textcolor{red}{2} N {O}_{\left(g\right)}$

Notice that the reaction produces $\textcolor{red}{2}$ moles of nitric oxide; the standard enthlapy of formation is always given for the formation of 1 mole of a substance from its elements in their most stable form.

This means that the enthalpy change for your reaction will be

2cancel("moles NO") * "90.25 kJ"/(1cancel("mole NO")) = "180.5 kJ"

The main equation you'll use is

$\Delta G = \Delta H - T \cdot \Delta S$, where

$\Delta G$ - the change in Gibbs free energy;
$\Delta H$ - the change in enthalpy;
$T$ - the temperature in Kelvin;
$\Delta S$ - the change in entropy.

To determine the change in entropy, use the standard entropies given to you

DeltaS_"rxn" = sum(n * S_"products"^0) - sum(m * S_"rectants"^0)

DeltaS_"rxn" = (2cancel("moles") * 211"J"/(cancel("mole") * K)) - (1cancel("mole") * 192"J"/(cancel("mole") * K) + 1cancel("mole") * 205"J"/(cancel("mole") * K))

$\Delta {S}_{\text{rxn" = "25 J/K}}$

Keep in mind, however, that the change in enthalpy is expressed in kJ, so convert the change in entropy to kJ per K

25cancel("J")/"K" * "1 kJ"/(1000cancel("J")) = 25 * 10^(-3)"kJ/K"

Now you have everything you need to solve for $\Delta G$. Plug your values into the main equation to get

DeltaG = "180.5 kJ" - (273.15 + 25)cancel("K") * 25 * 10^(-3)"kJ"/cancel("K")#

$\Delta G = \textcolor{g r e e n}{\text{+173 kJ}}$