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jok3333 [9.3K]
3 years ago
6

For this question, the "entropy term" refers to "-TΔS". Addition reactions are generally favorable at low temperatures because _

_______.
Chemistry
1 answer:
nata0808 [166]3 years ago
8 0

Answer:

Lowering the temperature typically reduces the significance of the decrease in entropy. That makes the Gibbs Free energy of the reaction more negative. As a result, the reaction becomes more favorable overall.  

Explanation:

In an addition reaction there's a decrease in the number of particles. Consider the hydrogenation of ethene as an example.

\rm H_2C\text{=}CH_2\; (g) + H_2\; (g) \stackrel{\text{Ni}^\ast}{\to} H_3C\text{-}CH_3\; (g).

When \rm H_2 is added to \rm H_2C\text{=}CH_2 (ethene) under heat and with the presence of a catalyst, \rm H_3C\text{-}CH3 (ethane) would be produced.

Note that on the left-hand side of the equation, there are two gaseous molecules. However, on the right-hand side there's only one gaseous molecule. That's a significant decrease in entropy. In other words, \Delta S < 0.

The equation for the change in Gibbs Free Energy for a particular reaction is:

\Delta G = \Delta H + (\underbrace{- T \, \Delta S}_{\text{entropy}\atop \text{term}}).

For a particular reaction, the more negative \Delta G is, the more spontaneous ("favorable") the reaction would be.

Since typically \Delta S < 0 for addition reactions, the "entropy term" of it would be positive. That's not very helpful if the reaction needs to be favorable.

T (absolute temperature) is always nonnegative. However, lowering the temperature could help bring the value of

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