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Determine \(\Delta G°\) for the reaction:

\[2 \text{NO}_2(g) \rightarrow \text{N}_2\text{O}_4(g)\]

at 25°C.

Given:
\[\Delta G°_f (\text{NO}_2) = 51.3 \, \text{kJ/mol}\]
\[\Delta G°_f (\text{N}_2\text{O}_4) = 97.8 \, \text{kJ/mol}\]

Answer :

The ∆G° for the reaction 2 NO₂(g) → N₂O₄(g) at 25°C is -4.8. This indicates that the reaction is spontaneous under standard conditions at 25°C.

To determine the ∆G° for the reaction 2 NO₂(g) → N₂O₄(g) at 25°C, we need to use the following formula:
∆G° = ∆G°f(products) - ∆G°f(reactants)

Given values:
NO₂ (g) has a ∆G°f of 51.3
N₂O₄ (g) has a ∆G°f of 97.8

First, we'll find the total ∆G°f for the reactants:
Since there are 2 moles of NO₂ in the reaction, we multiply its ∆G°f value by 2:
∆G°f(reactants) = 2 * 51.3 = 102.6

Next, we'll find the total ∆G°f for the products:
There is only 1 mole of N₂O₄, so its ∆G°f value remains unchanged:
∆G°f(products) = 1 * 97.8 = 97.8

Now, we can use the formula to find the ∆G° for the reaction:
∆G° = ∆G°f(products) - ∆G°f(reactants)
∆G° = 97.8 - 102.6
∆G° = -4.8

So, the ∆G° for the reaction 2 NO₂(g) → N₂O₄(g) at 25°C is -4.8. This indicates that the reaction is spontaneous under standard conditions at 25°C.

To know more about reactions refer here:

https://brainly.com/question/4482869#

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