Question

Study the effect of varying the percent excess air during the steady-flow combustion of hydrogen at a pressure of 1 atm. At what temperature will 97 percent of \(\mathrm{H}_{2}\) burn into \(\mathrm{H}_{2} \mathrm{O}\) ? Assume the equilibrium mixture consists of \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2}, \mathrm{O}_{2},\) and \(\mathrm{N}_{2}\).

Short answer

In this problem, we are asked to determine the temperature at which 97% of hydrogen (H2) will burn into water (H2O) in a steady-flow combustion process. The equilibrium mixture consists of H2O, H2, oxygen (O2), and nitrogen (N2). To solve this, we computed the moles of reactants and products in the equilibrium mixture and used the Gibbs free energy equation to calculate the temperature. The final answer is around 1481 K, which is the temperature at which 97% of H2 will burn into H2O.

Step-by-step solution

Determine the stoichiometry of the reaction

Write down the stoichiometric equation for the combustion of hydrogen with air. Air is assumed to be composed of 21% O2 and 79% N2 by volume. The complete combustion of hydrogen in the presence of air can be represented as follows: $$\mathrm{2H}_{2} + \mathrm{O}_{2} \rightarrow 2\mathrm{H}_{2} \mathrm{O}$$

Calculate the moles of reactants

Determine the moles of oxygen (O2) and nitrogen (N2) in the air, assuming a stoichiometric combustion of hydrogen (H2). At 97% combustion efficiency, 0.97 moles of H2 are turned into H2O. Given that H2 burns in a 2:1 ratio with O2, we have 0.485 moles of O2 required. Now, calculate the moles of N2 present in the air by using the air composition (79% N2 by volume). So for every mole of O2, there are 3.76 moles of N2 (since 79/21 = 3.76). Therefore, we have 0.485 * 3.76 = 1.8226 moles of N2 in the given volume of air.

Calculate the moles of products

Determine the moles of H2O, H2, O2, and N2 in the equilibrium mixture. Since 97% of H2 is burned, we have 0.97 * 2 = 1.94 moles of H2O formed. The remaining 3% of H2 has not reacted, which means we still have 0.03 * 2 = 0.06 moles of H2. The total moles of O2 in the equilibrium mixture are 0.485 - (1.94/2) = 0.015 moles, and that of N2 remains the same, 1.8226 moles.

Compute the temperature

To determine the temperature at which 97% of H2 burns into H2O, we'll use the Gibbs free energy equation for the combustion reaction. At equilibrium, the change in Gibbs free energy (\(\Delta G\)) is zero, such that: $$\Delta G = \Delta H - T \Delta S = 0$$ where \(\Delta H\) is the change in enthalpy, \(T\) is the temperature, and \(\Delta S\) is the change in entropy. Rearranging the equation, we get: $$T = \frac{\Delta H}{\Delta S}$$ Using data from steam tables or other sources, we can find the values of \(\Delta H\) and \(\Delta S\) for the reaction at standard temperature and pressure. For example, \(\Delta H\) for the combustion of hydrogen is around -241.8 kJ/mol, and \(\Delta S\) is around -163.3 J/mol·K. Plugging these values into the equation, we find: $$T = \frac{-241.8 \times 10^{3}\; \text{J/mol}}{-163.3\; \text{J/mol·K}} \approx 1481\; \text{K}$$ So, the temperature at which 97% of H2 will burn into H2O is around 1481 K.