question archive A) calculate the standard enthalpy change for the reaction 2A+B?2C+2D Use the following data: Substance ΔH?f (kJ/mol) A -245 B -387 C 223 D -523 B) For the reaction given in Part A, how much heat is absorbed when 3
A) calculate the standard enthalpy change for the reaction 2A+B?2C+2D Use the following data: Substance ΔH?f (kJ/mol) A -245 B -387 C 223 D -523 B) For the reaction given in Part A, how much heat is absorbed when 3
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A) calculate the standard enthalpy change for the reaction
2A+B?2C+2D
Use the following data:
| Substance | ΔH?f (kJ/mol) |
| A | -245 |
| B | -387 |
| C | 223 |
| D | -523 |
B)
For the reaction given in Part A, how much heat is absorbed when 3.70 mol of A reacts?
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Solution :-
A)Balanced reaction equation
2A + B ----- > 2C + 2D
Formula to calculate the enthalpy change of the reaction is as follows
Delta H reaction = sum of delta Hf product - sum of Delta Hf reactant
= [(C*2)+(D*2)] – [ (A*2)+(B*1)]
= [(223*2)+(-523*2)] – [ (-245*2)+(-387*1)]
= 277 kJ
Therefore the enthalpy change for the reaction is 277 kJ
B) 3.70 mol A
How much heat is absorbed
Form the balanced equation and the enthalpy change value we know that
2 mol A = 277 kJ
Therefore
3.70 mol A * 277 kJ / 2 mol A = 512.5 kJ
So the 3.70 mol A absorbs 512.5 kJ heat
