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Ammonium nitrate dissolves spontaneously and endothermally in water at room temperature. What can you deduce about the sign of ΔS for this solution process?

Short Answer

Expert verified
In this exercise, we are asked to deduce the sign of the change in entropy, ΔS, for the dissolution of ammonium nitrate in water. The dissolution is spontaneous and endothermic at room temperature. We know that for a process to be spontaneous, the Gibbs Free Energy change, ΔG, must be negative. Furthermore, the enthalpy change, ΔH, is positive since the process is endothermic. Analyzing the ΔG equation, we can deduce that the sign of ΔS must be positive, as this allows ΔG to be negative, which is necessary for a spontaneous process.

Step by step solution

01

Identify the given information

We know that the dissolution of ammonium nitrate in water is spontaneous and endothermic at room temperature. Spontaneous means that the process occurs without any external influence, and endothermic means that it absorbs heat from its surroundings.
02

Use the Gibbs Free Energy formula

The Gibbs Free Energy change, ΔG, can be used to determine the spontaneity of a process. The formula for Gibbs Free Energy change is: ΔG=ΔHTΔS where ΔG is the change in Gibbs Free Energy, ΔH is the change in enthalpy (heat), ΔS is the change in entropy, and T is the temperature in Kelvin.
03

Determine the sign of the enthalpy change

Since we are given that the dissolution is endothermic, this means that the process absorbs heat from its surroundings. Therefore, the enthalpy change, ΔH, is positive.
04

Analyze the spontaneity

For a process to be spontaneous, the Gibbs Free Energy change, ΔG, must be negative. Since we know that ΔH is positive in this case, we can deduce the sign of ΔS by analyzing the ΔG equation: - If ΔS is negative, the TΔS term would be positive, making ΔG positive and the process non-spontaneous. - If ΔS is positive, the TΔS term would be negative, and since ΔH is also positive, the overall ΔG could be negative and the process can be spontaneous.
05

Conclusion

Based on the given information, we can deduce that the sign of the change in entropy, ΔS, for the dissolution of ammonium nitrate in water must be positive. This is because it allows the Gibbs Free Energy change, ΔG, to be negative, which is necessary for a spontaneous process.

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Most popular questions from this chapter

Using S values from Appendix C, calculate ΔS values for the following reactions. In each case account for the sign of ΔS. (a) C2H4(g)+H2(g)C2H6(g) (b) N2O4(g)2NO2(g) (c) Be(OH)2(s)BeO(s)+H2O(g) (d) 2CH3OH(g)+3O2( g)2CO2( g)+4H2O(g)

For the isothermal expansion of a gas into a vacuum, ΔE=0,q=0, and w=0. (a) Is this a spontaneous

Indicate whether each statement is true or false. (a) If a system undergoes a reversible process, the entropy of the universe increases. (b) If a system undergoes a reversible process, the change in entropy of the system is exactly matched by an equal and opposite change in the entropy of the surroundings. (c) If a system undergoes a reversible process, the entropy change of the system must be zero. (d) Most spontaneous processes in nature are reversible.

For a certain chemical reaction, ΔHk=35.4 kJ and ΔSn=85.5 J/K. (a) Is the reaction exothermic or endothermic? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (c) Calculate ΔG for the reaction at 298 K. (d) Is the reaction spontaneous at 298 K under standard conditions?

Indicate whether cach statement is true or false. (a) The second law of thermodynamics says that entropy is conserved. (b) If the entropy of the system increases during a reversible process, the entropy change of the surroundings must decrease by the same amount. (c) In a certain spontaneous process the system undergoes an entropy change of 4.2 J/K; therefore, the entropy change of the surroundings must be 4.2 J/K.

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