Question

Determine at what temperature the reaction N2 3H2 -> 2NH3 will be product favored. delta H = -93 kJ/mol, delta S = -198 J/K mol?

The answer that I got is 469.7 K < T. Is this correct? Thank you for any help!

Answer 1

To determine thermodynamic favorability, we assume a reaction is thermodynamically favorable when `DeltaG<0`.

Recall,

`DeltaG = DeltaH - TDeltaS`

Given,

`DeltaH = (-93"kJ")/"mol"`

`DeltaS = (-198"J")/"K"`

`DeltaG = 0` (this is the threshold a process becomes thermodynamically favorable)

Hence,

`=> T = (DeltaH)/(DeltaS) approx 470"K"`

is the temperature at which the unfavorable entropy of this process is overcome.

The more we lower this temperature, the more the reaction will favor products.

Answer 2

This process is exothermic, so higher temperatures make it harder to work with in the industry.

We want the temperature for when the reaction will CEASE to be product-favored, and not when it will become spontaneous. That is not trivial, if I am to take your question as-written.

I get that `bb(T < "469.7 K")` (and not `T > "469.7 K"`). Above this temperature, the reaction STOPS being product favored, so we want the temperature to be lower to favor the products, not higher.

---

`K` varies with temperature, so we cannot simply say that `cancel(DeltaG)^(0) = DeltaG^@ + RTlncancel(Q)^(K)` and solve for `T`, because

`T = -(DeltaG^@)/(RlnK)`

which goes to infinity if `K = 1` as the question would be implying. Here are two ways to proceed.

METHOD 1

If `K = 1`, `DeltaG^@ = 0`, so

`0 = DeltaH^@ - TDeltaS^@`

As a result,

`T = (DeltaH^@)/(DeltaS^@) = (-"93 kJ/mol")/(-"0.198 kJ/mol"cdot"K"`

`=` `"469.7 K"`

We would have to recognize that since `DeltaH^@ < 0`, the higher temperature would disfavor the products. Therefore, `color(blue)(T_(max) = "469.7 K")` (not `T_(min)`).

METHOD 2

In this longer method, we could find `DeltaG^@`, and then use it to find `K_1`. Then, we could find `T` for which `K_2 = 1`. If this way works, we should get the same answer.

I would first find `DeltaG^@` at `"298.15 K"` (which is nonzero), and then find `K_1` at `"298.15 K"`:

`DeltaG^@ = DeltaH^@ - TDeltaS^@`

`= -"93 kJ"/"mol" - "298.15 K" cdot (-"0.198 kJ/mol"cdot"K")`

`= -"33.97 kJ/mol"`

The equilibrium constant at this temperature would then be:

`K_1 = e^(-(DeltaG^@)/(RT))`

`= "exp"[-(-"33.97 kJ/mol")/("0.008314472 kJ/mol"cdot"K" cdot "298.15 K")]`

`= e^(13.7)`

...Looks like it's VERY product-favored at `"298.15 K"` already... we can only go down in thermodynamic favorability for the forward reaction at higher temperatures for this exothermic process.

Then, I would use the van't Hoff equation to find what `T_2` temperature would lead to `K_2 = 1`, if we wanted to find the MAXIMUM temperature that it stays product-favored.

Again, since the reaction is exothermic, higher temperatures would STOP favoring the products and lower `K`, so we expect `T_2 > T_1`.

Assuming `DeltaH^@` stays constant in the temperature range:

`ln (K_2/K_1) = -(DeltaH^@)/R[1/T_2 - 1/T_1]`

`(Rln(1//K_1))/(-DeltaH^@) = 1/T_2 - 1/T_1`

`color(blue)(T_2) -= T_(max) = 1/((Rln(1//K_1))/(-DeltaH^@) + 1/T_1)`

`= 1/(("0.008314472 kJ/mol"cdot"K"cdotln(e^(-13.7)))/(-(-"93 kJ/mol")) + 1/"298.15 K")`

`=` `color(blue)("469.7 K")`

Above this temperature, the reactants become favored, so we want `T < "469.7 K"`.