Chemistry Labs

Problem 1

Fast-burning powders such as nitrocellulose are used in clay pigeon shooting cartridges; their burn rate is designed for lighter loads such as the 28 g clay pigeon load. Nitrocellulose is primarily composed of cellulose trinitrate, prepared by adding cellulose (empirical formula CX6HX7(OH)X3OX2\ce{C6H7(OH)3O2}) to a mixture of concentrated nitric and sulfuric acid, giving cellulose trinitrate (empirical formula CX6HX7(ONOX2)X3OX2\ce{C6H7(ONO2)3O2}) and one other product. The nitronium ion NOX2X+\ce{NO2+}, generated in the acid mixture, nitrates the alcohol groups. The nitric acid acts as a Lewis base and accepts a proton from sulfuric acid to form the intermediate HX2NOX3X+\ce{H2NO3+}, which then decomposes to the nitronium ion. (a) Write an equation for the reaction between cellulose and nitric acid to form cellulose trinitrate. (b) Write the equation for the formation of HX2NOX3X+\ce{H2NO3+} from nitric acid and sulfuric acid, and the equation for its decomposition to NOX2X+\ce{NO2+}. (c) Complete combustion of cellulose trinitrate produces carbon dioxide, water and a gaseous element. Write the equation for the complete combustion of CX6HX7NX3OX11\ce{C6H7N3O11}, and calculate the standard enthalpy of combustion using ΔfH∘(CO)=−110.5\Delta_f H^{\circ}(\ce{CO}) = -110.5, ΔfH∘(COX2)=−393.5\Delta_f H^{\circ}(\ce{CO2}) = -393.5, ΔfH∘(HX2O)=−285.8\Delta_f H^{\circ}(\ce{H2O}) = -285.8 and ΔfH∘(cellulose trinitrate)=−653.1\Delta_f H^{\circ}(\text{cellulose trinitrate}) = -653.1 kJ mol−1^{-1}. (d) When a shot is fired there is not enough time to react with oxygen from air; instead cellulose trinitrate decomposes entirely to gaseous products (no OX2\ce{O2} formed) and the products reach temperatures over 200 ∘C200\ ^{\circ}\mathrm{C}. Write the equation for this decomposition and calculate the total volume of gaseous products from 5.00 g of cellulose trinitrate at 200 ∘C200\ ^{\circ}\mathrm{C}, in m3^3.
Step 3 of 5: Balanced combustion
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CX6HX7NX3OX11+94 OX2→6 COX2+72 HX2O+32 NX2\ce{C6H7N3O11 + 9/4 O2 -> 6CO2 + 7/2 H2O + 3/2 N2}
Analysis

All carbon ends as COX2\ce{CO2}, hydrogen as HX2O\ce{H2O}, and nitrogen as the gaseous element NX2\ce{N2}; balancing oxygen then requires 94 OX2\tfrac{9}{4}\,\ce{O2}.