# Question 1fad2

Jun 5, 2015

For a reaction to be feasible the total entropy change of the system and surroundings must be positive.

${N}_{2 \left(g\right)} + 2 {O}_{2 \left(g\right)} \rightarrow 2 N {O}_{2 \left(g\right)}$

To work out the entropy change in the system we need to find the total entropy of the products and subtract this from the total entropy of the reactants.

$\Delta {S}_{s y s} = \Sigma \left[\text{entropy products"]-Sigma["entropy reactants}\right]$

You can get these data from standard tables:

${S}^{0} \left({N}_{2}\right) = 191.5 \text{J/K/mol}$

${S}^{0} \left({O}_{2}\right) = 205.3 \text{J/K/mol}$

${S}^{0} \left(N {O}_{2}\right) = 240.4 \text{J/K/mol}$

$\Delta {S}_{s y s} = \left(2 \times 240.4\right) - \left[191.5 + \left(2 \times 205.3\right)\right]$

$\Delta {S}_{s y s} = 480.8 - \left[602.1\right]$

$\Delta {S}_{s y s} = - 121.3 \text{J/K/mol}$

Now we need to work out the entropy change in the surroundings which would result if this reaction were to happen.

This is given by:

$\Delta {S}_{s u r r} = - \frac{\Delta H}{T}$

$\Delta {S}_{s u r r} = - \frac{66.4 \times {10}^{3}}{298} = - 222.8 \text{J/K/mol}$

Now we need to add these two together:

$\Delta {S}_{\text{total}} = \Delta {S}_{s y s} + \Delta {S}_{s u r r}$

$\Delta {S}_{\text{total}} = - 121.3 + \left(- 222.8\right)$

DeltaS_("total")=-344.1"J/K/mol"#

The negative value of $\Delta {S}_{\text{total}}$ tells us that the reaction cannot be spontaneous at this temperature.