Question

On April 16, 1947, a French ship containing 2300 tons (2.09 × 106 g) of ammonium nitrate caught fire in the port of Texas City, TX, and exploded. In this explosion, ammonium nitrate reacted according to NH4NO3 (s) → N2 (g) + ½ O2 (g) + 2 H2O (g) (a) Calculate the maximum energy lost in this accident that otherwise could have been used to do work, for example, through controlled explosions in mining enterprises. Even though it seems unrealistic for an explosion, assume that all products and reactants are provided and yielded at room temperature, respectively. (For NH4NO3, use: ΔHf° = -365.9 kJ mol-1 ; Sm° = 151.1 J mol-1 K-1 ) (b) Compare the numerical values for the enthalpy and entropy changes in the ammonium nitrate explosion and determine their percentage contribution to the maximum energy lost in the accident. (c) Determine whether the maximum energy would be larger or smaller if the products are yielded at a temperature higher than room temperature. (Hint: Consider the products to be ideal gases; Cp,m (H2) = 28.84 J mol-1 K-1 ; Cp,m (O2) = 29.38 J mol-1 K-1 ; Cp,m (H2O, g) = 33.59 J mol-1 K-1 )

4. (3 pts) On April 16, 1947, a French ship containing 2300 tons (2.09 * 106 g) of ammonium nitrate caught fire in the port of Texas City, TX, and exploded. In this explosion, ammonium nitrate reacted according to NH4NO3 (s) + N2(g) + 1/2O2(g) + 2 H2O (9) (a) Calculate the maximum energy lost in this accident that otherwise could have been used to do work, for example, through controlled explosions in mining enterprises. Even though it seems unrealistic for an explosion, assume that all products and reactants are provided and yielded at room temperature, respectively. (For NH4NO3, use: AH = -365.9 kJ mol-?; Smº = 151.1 J mol-1 K-) (b) Compare the numerical values for the enthalpy and entropy changes in the ammonium nitrate explosion and determine their percentage contribution to the maximum energy lost in the accident. (c) Determine whether the maximum energy would be larger or smaller if the products are yielded at a temperature higher than room temperature. (Hint: Consider the products to be ideal gases; Cp,m (H2) = 28.84 J mol-1 K-1; Com (O2) = 29.38 J mol K-1; Cp,m (H20,5) = 33.59 J molt K-)

Answer 1

Part (a).

The maximum energy lost (enthalpy) can be calculated as follows.

$\Delta$H_rxn = (2*$\Delta$H_H2O(g) + 1/2 * $\Delta$H_O2(g) + $\Delta$H_N2(g)) - ($\Delta$H_NH4NO3(s))

i.e. $\Delta$H^o_rxn = (2*-241.82 + 1/2 * 0 + 0) - (-365.9)

= -117.74 kJ/mol

The maximum energy lost (entropy) can be calculated as follows.

$\Delta$S^o_rxn = (2*S_H2O(g) + 1/2 * S_O2(g) + S_N2(g)) - (S_NH4NO3(s))

i.e. $\Delta$S_rxn = (2*188.72 + 1/2 * 205.03 + 191.50) - (151.1)

= 520.355 J/mol.K

Now, The maximum energy lost (free energy) can be calculated as follows.

$\Delta$G^o_rxn = $\Delta$H^o_rxn - T.$\Delta$S_rxn

= -117.74 kJ/mol - {(25+273) K * 0.520355 kJ/mol.K}

= -37.326 kJ/mol