Question

1. Calculate the change in enthalpy for this reaction: 2C(s) + H2(g) → C2H2(g) Given the following AHo -1299.5 kJ HP =-393.5 kJ HP -285.8 kJ CHI2(g) + (5/2)O2(g) → 2CO2(g) + H2O(1) C(s) + O2(g) → CO2(g) H2(g) + (12)O2(g) → H2O(1) A balloon contains 0.45 g of carbon dioxide and 0.6 grams of helium at 1.1 bar. If you compress the container to 80%ofits original volume, what will be the final pressure? Half a liter of water is cooled from 95°C to 50°C, and a 250 gram copper bar (originally at 25°C) is placed above it to capture the heat. Assuming that 40% of the heat escapes, what will be the final temperature of the copper bar? (Cp for copper is 0.385 J/gK) On the Supreme Scream ride at Knott's Berry Farm, riders are lifted to a height of 300 feet and then dropped. If there are 6 riders with an average weight of 140 lbs each, how much energy is needed to lift them to the top? 2. 3. 2 4. 5. A balloon contains 15 g of helium at room temperature and atmospheric pressure. 6. What is the change in enthalpy when a liter of water, originally at 25°C, is boiled and 7. Calculate ΔH for the decomposition of dinitrogen tetroxide to nitrogen dioxide at 8. Calculate AS for the above reaction at room temperature. How much work will be needed to compress it to half its original volume? vaporized? room temperature using the appendix. Then calculate ΔU without using the ΔU data. Calculate ΔH and ΔS for the above reaction at 55°C. 9.

Answer 1

2c(s) + H_2(g) rightarrow c_2H_2(g) Given, c_2H_2(g) + 5/2 o_2(g) = 2co_2(g) + H_2o(l) OH degree = -1299.5KJ Therefore C(s) + o_2(g) rightarrow co_2(g) oH degree = -393.5 KJ H_2(g) + 1/2 o2(g) rightarrow H_2o(l) Delta H degree = -285.8KJ Therefore 2x(s) + o2(g) rightarrow 2co_2(g) , Delta H degree = -787 KJ H_2(g) + 1/2 o2(g) rightarrow H_2o(l) , Delta H degree = -285.8 KJ 2 co_2(g) + H_2o(l) rightarrow C_2H_2(g) + 5/2o_2(g) Delta H degree = + 1299 . 5 kg 2c(S) + H_2(g) rightarrow C_2H_2(g) , Delta H degree = + 226.7 KJ Because The enthalpy change for this reaction is oH degree = + 226.7 KJ 0.45 gm co_2 = 0.45/44 = 0.0102 mole 0.6 gm He = 0.6/4 = 0.15 mole Total no. of moles n = (0.0102 + 0.15)mole = 0.1602 mole Let , the initial volume V & temperature T initial pressure = 1.1 bar Therefore PV = nRT or , 1.1V = 0.1602 RT or, T = 1.1V/0.1602 R Final volume V^1 = v times 80/100 = 0.8V let final pressure p^1 Therefore p^1v^1 = nRT or, p^1 = nRT/V^1 = 0.1602 times R/0.8V times 1.1V/0.1602 R