Question

the thermodynamics of ethanol
change in entropy for combustion
c₂h₅oh(l) + 3o₂(g) → 2co₂(g) + 3h₂o(g)
use the standard thermodynamic data in table 1 to calculate the entropy change associated with the combustion of ethanol, in j/mol·k.
thermodynamic table
compound | s° (j/mol k) | δh°f (kj/mol)
c(s, graphite) | 5.69 | 0.00
co₂(g) | 213.60 | -393.50
h₂(g) | 130.60 | 0.00
h₂o(g) | 188.8 | -241.82
o₂(g) | 205.00 | 0.00
c₂h₅oh(l) | 160.7 | -277.40
what change in entropy occurs when ethanol goes through combustion?
-217.9 j/mol k
37.0 j/mol k
-37.0 j/mol k
217.9 j/mol k

Explanation:

Step1: Recall the formula for entropy change

The formula for the standard entropy change ($\Delta S^{\circ}$) of a reaction is $\Delta S^{\circ}=\sum nS^{\circ}(\text{products})-\sum mS^{\circ}(\text{reactants})$, where $n$ and $m$ are the stoichiometric coefficients of the products and reactants respectively.

Step2: Identify the reactants and products with their stoichiometric coefficients

For the reaction $\ce{C_{2}H_{5}OH(l) + 3O_{2}(g)
ightarrow 2CO_{2}(g) + 3H_{2}O(g)}$:

• Reactants: $\ce{C_{2}H_{5}OH(l)}$ (coefficient = 1), $\ce{O_{2}(g)}$ (coefficient = 3)
• Products: $\ce{CO_{2}(g)}$ (coefficient = 2), $\ce{H_{2}O(g)}$ (coefficient = 3)

Step3: Find the standard entropy values from the table

• $S^{\circ}(\ce{C_{2}H_{5}OH(l)}) = 160.7\ \text{J/mol·K}$
• $S^{\circ}(\ce{O_{2}(g)}) = 205.00\ \text{J/mol·K}$
• $S^{\circ}(\ce{CO_{2}(g)}) = 213.60\ \text{J/mol·K}$
• $S^{\circ}(\ce{H_{2}O(g)}) = 188.8\ \text{J/mol·K}$

Step4: Calculate the sum of entropy of products

$\sum nS^{\circ}(\text{products})=2\times S^{\circ}(\ce{CO_{2}(g)}) + 3\times S^{\circ}(\ce{H_{2}O(g)})$
$=2\times213.60 + 3\times188.8$
$=2\times213.60+3\times188.8$
$=427.2 + 566.4$
$=993.6\ \text{J/mol·K}$

Step5: Calculate the sum of entropy of reactants

$\sum mS^{\circ}(\text{reactants})=1\times S^{\circ}(\ce{C_{2}H_{5}OH(l)})+3\times S^{\circ}(\ce{O_{2}(g)})$
$=160.7+3\times205.00$
$=160.7 + 615.00$
$=775.7\ \text{J/mol·K}$

Step6: Calculate the entropy change

$\Delta S^{\circ}=\sum nS^{\circ}(\text{products})-\sum mS^{\circ}(\text{reactants})$
$=993.6 - 775.7$
$=217.9\ \text{J/mol·K}$ Wait, no, wait. Wait, let's recalculate. Wait, no, I think I made a mistake. Wait, no:

Wait, recalculating the sum of products:
$2\times213.60 = 427.2$; $3\times188.8=566.4$; sum is $427.2 + 566.4 = 993.6$

Sum of reactants: $1\times160.7 + 3\times205.00 = 160.7 + 615 = 775.7$

Then $\Delta S^{\circ}=993.6 - 775.7 = 217.9$? But that's one of the options. Wait, but let's check again. Wait, no, wait the reaction is combustion of ethanol. Wait, maybe I mixed up? Wait, no, the formula is correct. Wait, the values:

Wait, $2\times213.60 = 427.2$; $3\times188.8 = 566.4$; total products: $427.2 + 566.4 = 993.6$

Reactants: $1\times160.7 + 3\times205 = 160.7 + 615 = 775.7$

$993.6 - 775.7 = 217.9$? But wait, the options have 217.9 J/mol K as an option. Wait, but let me check again. Wait, maybe I messed up the reaction. Wait, the reaction is $\ce{C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(g)}$. So yes, the coefficients are correct. So the calculation gives $\Delta S^{\circ}=217.9\ \text{J/mol·K}$. Wait, but let me check the arithmetic again:

$2\times213.60 = 427.2$

$3\times188.8 = 566.4$

Sum of products: $427.2 + 566.4 = 993.6$

Sum of reactants: $160.7 + 3\times205 = 160.7 + 615 = 775.7$

$993.6 - 775.7 = 217.9$

Yes, that's correct.

Answer:

217.9 J/mol K (the option with 217.9 J/mol K)