Question

Problem 1. A 35.6 g piece of aluminum metal rod is heated to 89.4 °C and then placed in an insulated container containing 51.2 g of water at 22.3 °C. Assuming no loss of water and a heat capacity of 19.82 J/K for the container, what is the final temperature of the system? You may assume that the container is always at the same temperature as that of the water. (HINT: qmetal + qcontainer + qwater-0) Problem 2. Calculate Δ formation n for the following unbalanced reaction using standard heats of C2H2()02(g) -> CO2(g) H20(g) Problem 3. Ammonia (NH3) reacts with oxygen at 30.4 °C in a 11.5 L vessel according the following unbalanced reaction equation. NH3(g) + 02(g) → N2(g) + H2O(g) The initial partial pressure of ammonia was 635.5 kPa and the initial partial pressure of oxygen was 508.1 kPa. (a) What is the initial total pressure within the vessel, in atm? (b) Which reactant is the limiting reactant? (c) How many moles of the excess reactant will remain? (d) How many moles of each of the products is produced? (e) What is the partial pressure of each gas, products and excess reactant, remaining in the reaction vessel, in kPa? (f) What is the final total pressure within the vessel, in atm?

Answer 1

Mass of metal rod, m35.6,g heat capacity of metal rod, C0.902J/g Initi al temperature of rod, T,xod -89.4 °C Let T bethe final temperature C heat of the metal ne myCT-) 35.6g x0.902 Jg ° Cx(T-89.4°C metal 32.1J°Cx(T-89.4 °c) Mass of water, mrar 51.2g heat capacity of metal rod, Ср.waar-4. 1 84 J/g °C Initi al temperature of water, T,woxe - 22.3°C Let be the final temperature heat of the metal, ggentera = PnAnterc,,water (T-. Tì.war 512gx4.184 Jg Cx(T-22.3°c) 2142JCx(T-22.3°c) heat capacity of container, Cy.cartniner- 19.82 J/K Initi al temperature of water, Ticataier - 22.3 oC Let be the final temperature heat of the container, qc.contairer.contirer conabrer 19.82JKx(T-22.3°c) Amount of heat rel eased by the hot metal rod is absorbed by the water and the container. Substitute the corresponding values from equations (1). (2) and (3)in the above equation 32 1JrCx(T-89.4°C)+214.2JCx(T-223 c)+1982JKx(T-223 c)- 32.17-2869.74 2 14.27-4776.66 +19.827-441.986 = 0 266.1278088.386 T 30.39 °C Ts 30.4 °C Therefore, the required soluti on is 30.4 °C