Question

Consider what happens when a sample of the explosive TNT (Section 8.8: "Chemistry Put to Work: Explosives and Alfred Nobel") is detonated. (a) Is the detonation a spontaneous process? (b) What is the sign of \(q\) for this process? (c) Can you determine whether \(w\) is positive, negative, or zero for the process? Explain. (d) Can you determine the sign of \(\Delta E\) for the process? Explain.

Short answer

The detonation of TNT is a spontaneous process, as its Gibbs free energy change (ΔG) is negative due to the release of a large amount of energy and increase in entropy. The sign of q for this process is negative, indicating heat is liberated from the system to its surroundings. The work (w) done during the detonation is negative, as it involves a rapid increase in volume. The change in internal energy (ΔE) is also negative, as both heat transfer (q) and work (w) are negative, implying a decrease in internal energy during the detonation of TNT.

Step-by-step solution

(a) Determine if the detonation is a spontaneous process.

For a process to be spontaneous, it must be thermodynamically favored, meaning its Gibbs free energy change (ΔG) should be negative. Explosives like TNT release a large amount of energy when detonated, and their enthalpy change (ΔH) is negative. The process is also accompanied by an increase in entropy (ΔS), as the solid TNT is transformed into gaseous products. Hence, both factors contribute to a negative ΔG, making the detonation of TNT a spontaneous process.

(b) Determine the sign of q for this process.

The heat transfer (q) is related to the enthalpy change (ΔH) of the process. As discussed earlier, the detonation of TNT is an exothermic process, meaning it releases heat into its surroundings. Therefore, the sign of q for this process is negative, as heat is liberated from the system to its surroundings.

(c) Determine if w is positive, negative, or zero for the process.

The work (w) done during the process depends on the change in volume, pressure, and the nature of the process. In the case of a detonation, the explosive nature of the process leads to a rapid and significant increase in volume (from solid TNT to gaseous products). Since the work done during an expansion is negative (w = -PΔV), we can conclude that the work done during the detonation of TNT is negative.

(d) Determine the sign of ΔE for the process.

The change in internal energy (ΔE) is given by the first law of thermodynamics: ΔE = q + w, where q is the heat transfer, and w is the work done. We know from our analysis that q is negative (heat is transferred from the system) and w is also negative (negative work done due to expansion). So, when we add these two negative values, the overall ΔE will be negative. This means the detonation of TNT results in a decrease in internal energy.