Assume that 2 moles of water are formed according to the following reaction at constant pressure \((101.3 \mathrm{kPa})\) and constant temnerature $(298 \mathrm{~K});$ $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) Calculate the pressure-volume work for this reaction. (b) Calculate \(\Delta E\) for the reaction using your answer to (a).

First, we need to find the initial and final moles of gas in the reaction. Initially, we have 2 moles of hydrogen gas and 1 mole of oxygen gas. So, initial moles of gas: \(n_{initial} = 2 + 1 = 3\) moles After the reaction, 2 moles of water (liquid) are formed, and there are no moles of gas left in the reaction. So, final moles of gas: \(n_{final} = 0\) moles Now, we'll use the Ideal Gas Law, with the given temperature \(T = 298 \mathrm{K}\) and pressure \(P = 101.3\,\mathrm{kPa}\), to find the initial and final volumes: $$PV = nRT$$ Where \(R\) is the gas constant in \(\mathrm{kPa \cdot L/mol\cdot K}\) which is equal to $8.314 \,\mathrm{J/mol\cdot K} \div 1000 = 0.008314\,\mathrm{kPa\cdot L/mol\cdot K}\) Initial volume: $$V_{initial} = \frac{n_{initial}RT}{P} = \frac{3(0.008314)(298)}{101.3} = 0.0737 \, \mathrm{L}$$ Final volume (V_final) is 0 since there are no moles of gas left in the reaction \((n_{final} = 0)\). The volume change (ΔV) during the reaction: $$\Delta V = V_{final} - V_{initial} = 0 - 0.0737 = -0.0737 \, \mathrm{L}$$

The pressure-volume work for a constant pressure process is given by: $$W = -P\Delta V$$ Now, plugging the given pressure and calculated volume change: $$W = -(101.3)(-0.0737) = 7.47 \, \mathrm{kJ}$$ So, the pressure-volume work for this reaction is 7.47 kJ.

Since we're given the pressure-volume work (W) and asked to find the change in internal energy (ΔE), we can use the following equation: $$ΔE = q + W$$ However, we're not given the heat transferred (q) directly. But we can use the following fact: for any reaction at constant temperature and pressure, the heat transferred is equivalent to the change in enthalpy (ΔH) of the reaction. For the formation of 2 moles of water from hydrogen and oxygen gases, the change in enthalpy is: $$ΔH = -571.6 \, \mathrm{kJ/mol}$$ Since 2 moles of water are formed, the total change in enthalpy is: $$q = 2 \times (-571.6) = -1143.2 \, \mathrm{kJ}$$ Now we can find the change in internal energy using the relation: $$ΔE = q + W = -1143.2 + 7.47 = -1135.73 \, \mathrm{kJ}$$ Therefore, the change in internal energy (ΔE) for the reaction is -1135.73 kJ.