Question

Consider the following reaction: $$ 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=-572 \mathrm{~kJ} $$ a. How much heat is evolved for the production of \(1.00 \mathrm{~mol}\) \(\mathrm{H}_{2} \mathrm{O}(l) ?\) b. How much heat is evolved when \(4.03 \mathrm{~g}\) hydrogen is reacted with excess oxygen? c. How much heat is evolved when \(186 \mathrm{~g}\) oxygen is reacted with excess hydrogen? d. The total volume of hydrogen gas needed to fill the Hindenburg was \(2.0 \times 10^{8} \mathrm{~L}\) at \(1.0 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\). How much heat was evolved when the Hindenburg exploded, assuming all of the hydrogen reacted?

Short answer

The heat evolved for the production of 1.00 mol H2O is -286 kJ. When 4.03 g hydrogen is reacted with excess oxygen, the heat evolved is -572 kJ. When 186 g oxygen is reacted with excess hydrogen, the heat evolved is -3325.25 kJ. The heat evolved when the Hindenburg exploded, assuming all of the hydrogen reacted, was approximately -2.33 x 10^9 kJ.

Step-by-step solution

Use stoichiometry to find the moles of H2O formed

Since the balanced chemical equation shows that 2 moles of H2 react with 1 mole of O2 to form 2 moles of H2O, it means that for the production of 1 mole of H2O, the reaction consumes 1 mole of H2 and 0.5 moles of O2.

Calculate heat evolved

As the enthalpy change given (-572 kJ) is for the reaction of 2 moles of H2 with 1 mole of O2, we need to find the enthalpy change for the formation of 1 mole of H2O. Enthalpy change for 1 mol H2O = \(\frac{-572\,\mathrm{kJ}}{2}\) = \( -286\,\mathrm{kJ}\) Answer: The heat evolved for the production of 1.00 mol H2O is -286 kJ. #b. Heat evolved for 4.03 g of hydrogen reacting#

Convert grams to moles

To find the moles of hydrogen gas, we use the molar mass of hydrogen, which is 2.016 g/mol. Moles of H2 = \(\frac{4.03\,\mathrm{g}}{2.016\,\mathrm{g/mol}} \approx 2.00\,\mathrm{mol}\)

Calculate heat evolved

Since 2 moles of H2 react with 1 mole of O2 to produce 2 moles of H2O and release -572 kJ, for the reaction of 2 moles of hydrogen gas, the heat evolved will be -572 kJ. Answer: The heat evolved when 4.03 g hydrogen is reacted with excess oxygen is -572 kJ. #c. Heat evolved for 186 g of oxygen reacting#

Convert grams to moles

To find the moles of oxygen gas, we use the molar mass of oxygen, which is 32.00 g/mol. Moles of O2 = \(\frac{186\,\mathrm{g}}{32.00\,\mathrm{g/mol}} = 5.8125\,\mathrm{mol}\)

Calculate heat evolved

From the balanced chemical equation, 1 mole of O2 reacts with 2 moles of H2 to produce 2 moles of H2O and release -572 kJ of heat. Therefore, for the reaction of 5.8125 moles of oxygen gas: Heat evolved = \((5.8125\,\mathrm{mol\,O_{2}}) \times (-572\,\mathrm{kJ/mol\,O_{2}}) = -3325.25\,\mathrm{kJ}\) Answer: The heat evolved when 186 g oxygen is reacted with excess hydrogen is -3325.25 kJ. #d. Heat evolved for the explosion of the Hindenburg#

Determine moles of hydrogen gas

We are given the volume, pressure, and temperature of hydrogen gas in the Hindenburg. We will use the ideal gas law to find the moles of hydrogen gas: PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is temperature in Kelvin. Rearrange the equation to solve for n: n = \(\frac{PV}{RT}\) Given that P = 1.0 atm, V = \(2.0\times10^{8}\,\mathrm{L}\), R = 0.0821 \(L\cdot atm\,K^{-1}\ mol^{-1}\), and T = 298 K (25 degrees Celsius converted to Kelvin): n = \(\frac{(1.0\,\mathrm{atm})(2.0\times10^{8}\,\mathrm{L})}{(0.0821\,\mathrm{L\,atm\,K^{-1}\,mol^{-1}})(298\,\mathrm{K})} = 8.15\times10^{6}\,\mathrm{mol}\) The Hindenburg contains 8.15 x 10^6 moles of hydrogen.

Calculate heat evolved

Since the chemical equation shows that 2 moles of H2 react with 1 mole of O2 to produce 2 moles of H2O and release -572 kJ of heat, we can calculate the heat evolved in the explosion by: Heat evolved = \((8.15\times10^{6}\,\mathrm{mol\,H_{2}}) \times \frac{-572\,\mathrm{kJ}}{2\,\mathrm{mol\,H_{2}}} = -2.33\times10^{9}\,\mathrm{kJ}\) Answer: The heat evolved when the Hindenburg exploded, assuming all of the hydrogen reacted, was approximately -2.33 x 10^9 kJ.

Key concepts

Stoichiometry

Stoichiometry is the art of balancing chemical equations and allows us to calculate the amounts of reactants or products involved in a chemical reaction. In the given exercise, stoichiometry is applied to understand how much heat is evolved when different amounts of hydrogen and oxygen are reacted to form water.

For example, the equation \(2 \text{H}_2(g) + \text{O}_2(g) \rightarrow 2 \text{H}_2\text{O}(l)\) shows that 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of liquid water. This means that for every mole of oxygen used, two moles of hydrogen are required, demonstrating a specific ratio required for the reaction to occur fully.

To calculate the heat evolved in the exercise, it's crucial to first understand the role of stoichiometry, which ensures the correct proportions of reactants are used to determine the amount of energy released or absorbed in a reaction.

Ideal Gas Law

The ideal gas law is a key principle in chemistry, particularly used to relate the properties of gases. It is expressed as \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the amount of substance (in moles), \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin.

This law is crucial in solving part d of the exercise which involves determining the moles of hydrogen gas in the Hindenburg. Given the volume of hydrogen was \(2.0\times10^8\) L at 1 atm and 25°C, converting this to moles helps us understand how much hydrogen was present and thus how much heat was released upon reaction.

By rearranging the ideal gas law formula to \(n = \frac{PV}{RT}\), we find that the Hindenburg contained \(8.15\times10^6\) moles of hydrogen gas. Understanding the ideal gas law allows us to accurately predict the behavior of gases under different conditions, making it indispensable in chemical calculations.

Molar Mass

Molar mass is the weight of one mole of a substance, measured in grams per mole (g/mol). It serves as a bridge between the mass of a material and the number of moles, which is pivotal in several parts of the exercise, particularly when converting grams to moles.

For hydrogen, the molar mass is approximately 2.016 g/mol, while for oxygen, it is 32.00 g/mol. These values allow us to calculate the moles of substances like hydrogen and oxygen from their given masses. For instance, in part b, 4.03 g of hydrogen is converted to moles as \(\frac{4.03 \,\text{g}}{2.016 \,\text{g/mol}} \approx 2.00 \,\text{mol}\).

Similarly, in part c, 186 g of oxygen is converted using its molar mass. Knowing the molar mass not only helps in carrying out these conversions, but it also allows one to determine the proportion of reactants in a given chemical reaction, critical for stoichiometric calculations.

Heat Evolution

Heat evolution refers to the heat released or absorbed during a chemical reaction. It's a way to quantify the energy change that occurs and is usually tied to the concept of enthalpy, denoted by \(\Delta H\).

In this exercise, the enthalpy change for the reaction where \(2 \text{H}_2\) and \(\text{O}_2\) form \(2 \text{H}_2\text{O}\) is given as \(-572\,\text{kJ}\). A negative sign indicates that the reaction is exothermic, meaning it releases heat.

Part a of the task involved determining how much heat is evolved when 1 mol of water forms, calculated as \(-286\,\text{kJ}\) by halving the total heat of the reaction. Understanding heat evolution is key in chemical engineering and environmental chemistry as it impacts how reactions are scaled up or controlled in real-world applications.