Nitroglycerine, \(\mathrm{C}_{3} \mathrm{H}_{5}\left(\mathrm{NO}_{3}\right)_{3}(l)\), is a powerful explosive used in rock blasting when roads are created. When ignited, it produces water, nitrogen, carbon dioxide, and oxygen. Detonation of one mole of nitroglycerine liberates \(5725 \mathrm{~kJ}\) of heat. (a) Write a balanced thermochemical equation for the reaction for the detonation of four moles of nitroglycerine. (b) What is \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{C}_{3} \mathrm{H}_{5}\left(\mathrm{NO}_{3}\right)_{3}(l) ?\)

Short Answer

Expert verified
a) The balanced thermochemical equation for the detonation of nitroglycerin is: $$4\,\text{C}_{3}\text{H}_{5}\text{N}_{3}\text{O}_{9}(l) \rightarrow 6\,\text{CO}_{2}(g) + 10\,\text{H}_{2}\text{O}(l) + \frac{1}{2}\,\text{O}_{2}(g) + 6\,\text{N}_{2}(g) \hspace{.5cm} \Delta H=-22900\,\text{kJ}$$ b) The standard enthalpy of formation (\(\Delta H_{\mathrm{f}}^{\circ}\)) for nitroglycerine is \(-5725\,\text{kJ/mol}\).

Step by step solution

01

Writing the Unbalanced Equation

First, let's write the unbalanced equation for the reaction of nitroglycerine: $$\text{C}_{3}\text{H}_{5}\text{N}_{3}\text{O}_{9}(l) \rightarrow \text{CO}_{2}(g) + \text{H}_{2}\text{O}(l) + \text{O}_{2}(g) + \text{N}_{2}(g)$$
02

Balancing the Equation

Now let's balance the equation. The balanced chemical equation is: $$4\,\text{C}_{3}\text{H}_{5}\text{N}_{3}\text{O}_{9}(l) \rightarrow 6\,\text{CO}_{2}(g) + 10\,\text{H}_{2}\text{O}(l) + \frac{1}{2}\,\text{O}_{2}(g) + 6\,\text{N}_{2}(g)$$ Here we have balanced the number of carbon, hydrogen, nitrogen, and oxygen atoms on both sides of the equation.
03

from a) Balanced Thermochemical Equation for the Detonation with 4 moles of nitroglycerine

Now, as we're asked to write a balanced thermochemical equation for the reaction, let's include the heat liberated for the 4 moles of nitroglycerine detonation. $$4\,\text{C}_{3}\text{H}_{5}\text{N}_{3}\text{O}_{9}(l) \rightarrow 6\,\text{CO}_{2}(g) + 10\,\text{H}_{2}\text{O}(l) + \frac{1}{2}\,\text{O}_{2}(g) + 6\,\text{N}_{2}(g) \hspace{.5cm} \Delta H=-4\cdot5725\,\text{kJ}$$ So that the answer for part (a) will be: $$4\,\text{C}_{3}\text{H}_{5}\text{N}_{3}\text{O}_{9}(l) \rightarrow 6\,\text{CO}_{2}(g) + 10\,\text{H}_{2}\text{O}(l) + \frac{1}{2}\,\text{O}_{2}(g) + 6\,\text{N}_{2}(g) \hspace{.5cm} \Delta H=-22900\,\text{kJ}$$
04

Calculating the \(\Delta H_{\mathrm{f}}^{\circ}\) for nitroglycerine

Now let's calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for 1 mole of nitroglycerine, given the released heat for 4 moles of nitroglycerin detonation as \(-22900\,\text{kJ}\). $$\Delta H_{\mathrm{f}}^{\circ} (\text{C}_{3}\text{H}_{5}\text{N}_{3}\text{O}_{9}) = \frac{-22900\,\text{kJ}}{4\,\text{moles}} = -5725\,\text{kJ/mol}$$ The answer for part (b) is \(\Delta H_{\mathrm{f}}^{\circ}(\text{C}_{3}\text{H}_{5}\text{N}_{3}\text{O}_{9})=-5725\,\text{kJ/mol}\).

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Most popular questions from this chapter

Write thermochemical equations for the formation of one mole of the following compounds from the elements in their stable states at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\). (a) acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\) (b) nitrogen dioxide \((g)\) (c) lead(II) bromide (s) (d) phosphorus pentachloride \((g)\)

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