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For the reaction A + B → C + D, ΔH° = +40 kJ and ΔS° = +50 J/K. Therefore, the reaction under standard conditions is: A) spontaneous only at temperatures between 10 K and 800 K. B) spontaneous at temperatures greater than 800 K. C) nonspontaneous at all temperatures. D) spontaneous at all temperatures. E) spontaneous at temperatures less than 10 K.

User EmKay
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2 Answers

7 votes

Final answer:

For the reaction A + B → C + D, with ΔH° = +40 kJ and ΔS° = +50 J/K, the reaction is spontaneous at temperatures greater than 800 K as the positive ΔH° and ΔS° indicates that an increase in temperature will lead to a negative Gibbs free energy change, indicating spontaneity.

Step-by-step explanation:

The question is determining the spontaneity of the reaction A + B → C + D with a given enthalpy change (ΔH°) and entropy change (ΔS°). Spontaneity is assessed by the Gibbs free energy change (ΔG), which is calculated by the equation ΔG = ΔH° - TΔS°, where T is the temperature in Kelvin.

Given that ΔH° = +40 kJ and ΔS° = +50 J/K, we need to determine the temperature at which ΔG becomes negative, as this indicates spontaneity. Since ΔH° is positive and ΔS° is also positive, the spontaneity depends on the temperature. Here we can use the formula:

ΔG = ΔH° - TΔS°

Spontaneity requires that ΔG < 0. Therefore, we solve for T:

0 > ΔH° - TΔS°
ΔH° < TΔS°
T > ΔH°/ΔS°
T > 40,000 J / 50 J/K
T > 800 K

Thus, the reaction in question is spontaneous at temperatures greater than 800 K, because it is only above this temperature that ΔG turns negative, making choice (B) the correct answer.

User Charles Ingalls
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4.9k points
7 votes

Answer:

spontaneous at temperatures greater than 800 K.

Step-by-step explanation:

From the thermodynamic equation ,

ΔG = ΔH - TΔS

For spontaneous reaction ΔG = - ve

ΔH - TΔS < 0

TΔS > ΔH

T > ΔH / ΔS

> 40000 / 50

> 800 K

At temperature greater than 800K , reaction will be spontaneous.

User Maikel Willemse
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5.3k points