Two moles of O2(g) are used to heat one mole of Q(s) to produce 1 mole of a gaseous compound in a small closed furnace. What is the ratio of the final pressure at 616 K to the initial pressure at 308 K? Assume all reactants are converted to products.

Explanation

The balanced chemical equation for the reaction is:Q(s) + O2(g) → QO2(g)According to the problem, 2 moles of O2(g) are used to produce 1 mole of QO2(g). Therefore, the number of moles of Q(s) used must also be 1. At the beginning of the reaction, the pressure in the container is due to the O2 gas only, so we can use the ideal gas law to calculate the initial pressure:As P=nRT/VPinitial = (2 mol)× R ×(308 K)/VAfter the reaction, the pressure in the container is due to the QO2 gas only, so we can use the ideal gas law again to calculate the final pressure:Pfinal = (1 mol)× R ×(616 K)/VTherefore, the ratio of final pressure to initial pressure is:Pfinal / Pinitial = (1 mol)× R ×(616 K)/V ÷ (2 mol)× R ×(308 K)/VRearranging the equation616/616 = 1So the correct answer is C) 1.