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Answer 1 · 2015-11-07T13:36:24

$+ 133 kJ/mol$ $+"133 kJ/mol"$

Explanation:

The idea here is that you need to use the change in Gibbs free energy, $Δ G^{\circ}$ $DeltaG^@$ , and the change in entropy, $Δ S^{\circ}$ $DeltaS^@$ , of the reaction at ${161.0}^{\circ} C$ $161.0^@"C"$ to determine the change in enthalpy of the reaction, $Δ H^{\circ}$ $DeltaH^@$ .

Assuming that all the species that take part in the reaction are present at standard-state conditions, i.e. a pressure of $1 atm$ $"1 atm"$ , you can use $Δ G^{\circ}$ $DeltaG^@$ , $Δ S^{\circ}$ $DeltaS^@$ , and $Δ H^{\circ}$ $DeltaH^@$ , instead of $Δ G$ $DeltaG$ , $Δ S$ $DeltaS$ , and $Δ H$ $DeltaH$ . This is just a notation thing, it doesn't affect what's going on.

So, the equation that establishes a relationship between a reaction's change in enthalpy, its change in entropy, and the temperature at which it takes place looks like this

$Δ G^{\circ} = Δ H^{\circ} - T \cdot Δ S^{\circ}$ $color(blue)(DeltaG^@ = DeltaH^@ - T * DeltaS^@)$

Here $Δ G^{\circ}$ $DeltaG^@$ is called the change in Gibbs free energy, and it tells you whether or not a particular reaction is spontaneous or non-spontaneous.

In essence, what that equation tells you is what is the driving force behind a particular reaction. In order for a reaction to be spontaneous, you need to have $Δ G^{\circ} < 0$ $DeltaG^@<0$ .

This means that you can have

$Δ H < 0$ $DeltaH<0$ , $Δ S > 0 \to$ $DeltaS>0 ->$ spontaneous at any temperature
$Δ H < 0$ $DeltaH<0$ , $Δ S < 0 \to$ $DeltaS<0 ->$ spontaneous at a certain temperature range
$Δ H > 0$ $DeltaH>0$ , $Δ S < 0 \to$ $DeltaS<0 ->$ non-spontaneous regardless of temperature
$Δ H > 0$ $DeltaH>0$ , $Δ S > 0 \to$ $DeltaS>0 ->$ spontaneous at a certain temperature range

Notice that the reaction is non-spontaneous at ${161.0}^{\circ} C$ $161.0^@"C"$ , since it has $Δ G^{\circ} > 0$ $DeltaG^@>0$ . Plug in your values and find its enthalpy change of reaction - do not forget to convert the temperature to Kelvin!

$Δ H^{\circ} = Δ G^{\circ} + T \cdot Δ S^{\circ}$ $DeltaH^@ = DeltaG^@ + T * DeltaS^@$

$Δ H^{\circ} = 22.20 kJ/mol + (273.15 + 161.0) K \cdot 805.89 \frac{J}{mol \cdot K}$ $DeltaH^@ = "22.20 kJ/mol" + (273.15 + 161.0)color(red)(cancel(color(black)("K"))) * 805.89"J"/("mol" * color(red)(cancel(color(black)("K"))))$

$Δ H^{\circ} = 22.20 kJ/mol + 3499877 J/mol$ $DeltaH^@ = "22.20 kJ/mol" + 3499877"J/mol"$

$Δ H^{\circ} = 22.20 kJ/mol + 349.9 kJ/mol = + 372.1 kJ/mol$ $DeltaH^@ = "22.20 kJ/mol" + "349.9 kJ/mol" = +"372.1 kJ/mol"$

So, at ${161.0}^{\circ} C$ $161.0^@"C"$ , the change in entropy is positive and the change in enthalpy is positive as well, which means that the temperature actually determines whether or not the reaction is spontaneous.

Plug in your values and find the change in Gibbs free energy at ${24.0}^{\circ} C$ $24.0^@"C"$

$Δ G^{\circ} = 372.1 kJ/mol - (273.15 + 24) K \cdot 805.89 \frac{J}{mol \cdot K}$ $DeltaG^@ = "372.1 kJ/mol" - (273.15 + 24)color(red)(cancel(color(black)("K"))) * 805.89"J"/("mol" * color(red)(cancel(color(black)("K"))))$

$Δ G^{\circ} = 372.1 kJ/mol - 239470 J/mol$ $DeltaG^@ = "372.1 kJ/mol" - "239470 J/mol"$

$Δ G^{\circ} = 372.1 kJ/mol - 239.5 kJ/mol = + 133 kJ/mol$ $DeltaG^@ = "372.1 kJ/mol" - "239.5 kJ/mol" = +color(green)("133 kJ/mol")$

Once again, the reaction comes out to be non-spontaneous. In fact, $Δ G^{\circ}$ $DeltaG^@$ increased compared with what the reaction had at ${161.0}^{\circ} C$ $161.0^@"C"$ , which means that it actually became less favorable as temperature decreased.

By the same logic, it will eventually become spontaneous as temperature increases. To test that, take the limit value $Δ G^{\circ} = 0$ $DeltaG^@ = 0$ and find the minimum temperature that will allow for that to happen

$0 = Δ H^{\circ} - T \cdot Δ S^{\circ} \Rightarrow Δ H^{\circ} = T \cdot Δ S^{\circ}$ $0 = DeltaH^@ - T * DeltaS^@ implies DeltaH^@ = T * DeltaS^@$

Therefore, you have

$T = \frac{Δ H^{\circ}}{Δ S^{\circ}}$ $T = (DeltaH^@)/(DeltaS^@)$

$T = \frac{372.1 \frac{kJ}{mol}}{0.80589 \frac{kJ}{mol \cdot K}} = 461.7 K$ $T = (372.1color(red)(cancel(color(black)("kJ")))/color(red)(cancel(color(black)("mol"))))/(0.80589color(red)(cancel(color(black)("kJ")))/(color(red)(cancel(color(black)("mol"))) * "K")) = "461.7 K"$

So, in order for this reaction to be spontaneous, you need to have

$T > 462 K$ $T > "462 K" " "$ , or $T > 189^{\circ} C$ $" "T> 189^@"C"$

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