THERMODYNAMICS: Given the following hypothetical thermochemical equations 2QJ3 (g) + 3Q2X (g) → 4Q2 (g) + 3J2X (1) AH = -1,070 kJ 4QJ3 (g) + 3X2 (g) –→ 202 (g) + 6J2X (1) AH = -1,580.2 kJ Calculate the value of AH (in kJ) for the reaction: Q2 (g) + ½ X2 (g) → Q2X (g (Round off the final answer to ONE decimal place. Do not include unit.)

You are given the following related reactions and the corresponding enthalpy changes. 1. 2. 2CO (g) N + O + O 2(g) 2CO (g) 0₂ (8) 2(g) + 2NO (g) → 2CO Use are to use the above reactions to determine AH for the following reaction. A AH = -566.0 kJ 2(g) → 2NOAH = 180.6 kJ → 2CO + N 2(g) 2(g) In the last problem, you reversed the second reaction, and wrote the new re- action and enthalpy. Now, combine the two reactions, canceling 2(g). Write the combined reaction. Then, add the enthalpies.

Given the following thermochemical equations, 2Cu (s) + S (s) · → Cu₂S (s) AH° = -79.5 kJ S (s) + O₂ (g) → SO₂ (g) AH° = −297 kJ Cu₂ S (s) +20₂ (g) → 2CuO (s) + SO₂ (g) AH° = -527.5 kJ calculate the standard enthalpy of formation (in kJ mole¹¹) of CuO (s).

4 CH3NH2 (g) + 2 H2O (l) ➝ CO2 (g) + 3 CH4 (g) + 4 NH3 (g)     ΔHº = –138.96 kJ• Calculate the overall change in enthalpy when 30.00 grams of CH3NH2 (g) reacts with excess H2O (l).   [Be sure to include the correct sign.]