Just filling in the blanks

Just filling in the blanks


Experimental Data and Results For each problem, find the ΔH for the target reaction below, given the following step reactions and subsequent ΔH values. Show step reactions, enthalpy values and your answer in the boxes provided. For full credit, ALL selections should be indicated, arrows should line up, and work shown in the manner indicated in the lecture video.   1. Target Reaction: A2B4 (l) + B2 (g) → 2 AB3 (g) Step Reactions: 1. A2B4 (l) + CB4O (l) → CB2O (g) + A2 (g) + 3 B2 (g)ΔH = - 37 kJ  2. A2 (g) + 3 B2 (g) → 2 AB3 (g)ΔH = - 46 kJ 3. CB4O (l) → CB2O (g) + B2 (g) ΔH = - 65 kJ Reaction # How modified Modified Step Reactions modified ?H Value 1 Select: forward or reverse Select: as is / multiplied by / divided by 2 Select: forward or reverse Select: as in / multiplied by / divided by 3 Select: forward or reverse Select: as in / multiplied by / divided by Answer (∆H) = 0. Target Reaction: XY5 (g) → XY3 (g) + Y2 (g) Step Reactions: 1. X4 (s) + 6 Y2 (g) → 4 XY3 (g)               ΔH = - 2439 kJ  2.  4 XY5 (g) → X4 (s) + 10 Y2 (g) ΔH =   3438 kJ Reaction # How modified Modified Step Reactions modified ?H Value 1 Select: forward or reverse Select: as is / multiplied by / divided by 2 Select: forward or reverse Select: as is / multiplied by / divided by Answer (∆H) = 0. Target Reaction: 2 XZ2 (g) + H2Z (g) → X 2H2 (g) + 52 Z2 (g) Step Reactions: 1. X2H2 (g) + 2 H2 (g) → X2H6 (g) ΔH =  - 94.5 kJ    2.    H2Z (g) → H2 (g) + ½ Z2 (g)               ΔH =    71.2 kJ        3.    2 X2H6 (g) + 7 Z2 (g) → 4 XZ2 (g) + 6 H2Z (g)      ΔH =  - 566 kJ        Reaction # How modified Modified Step Reactions modified ?H Value 1 Select: forward or reverse Select: as is / multiplied by / divided by 2 Select: forward or reverse Select: as is / multiplied by / divided by 3 Select: forward or reverse Select: as is / multiplied by / divided by Answer (∆H) = Additional Modification of ?H Needed? Select: Yes or no Select: As is / multiplied by / divided by