Part A
Calculate the standard enthalpy change for the reaction
2A+B --->2C +2D
Use the following data:
Substance Delta H(kJ/mol)
A -263
B -391
C +203
D -523
Express your answer in kilojoules.
MY ANSWER FOR PART A: Delta H =277 kJ
PART B----THIS ONE I NEED HELP ON
Express your answer numerically in kilojoules
For the reaction given in Part A, how much heat is absorbed when 3.80 mol of A reacts?
See your post above.
To find the heat absorbed (ΔH) when a certain amount of substance reacts, you need to use the stoichiometry of the reaction and the given enthalpy change values.
In Part A, you calculated the standard enthalpy change for the reaction as ΔH = 277 kJ. This value represents the enthalpy change per mole of reaction, so it can be used to calculate the heat absorbed for any given amount of substance.
Now, let's move on to Part B.
Given:
Amount of substance A = 3.80 mol
To find the heat absorbed when 3.80 mol of A reacts, you can use the stoichiometry of the reaction. From the balanced equation:
2A + B → 2C + 2D
You can see that 2 moles of A react to form 2 moles of C and 2 moles of D. Therefore, the molar ratio between A and ΔH is 2:ΔH, or simply 2ΔH.
To calculate the heat absorbed for the given amount of substance A, you can use the formula:
Heat = (Amount of substance) × (ΔH per mole)
Plugging in the values:
Heat = 3.80 mol × (2ΔH)
Since ΔH was previously calculated as 277 kJ, we can substitute it into the formula:
Heat = 3.80 mol × (2 × 277 kJ)
Calculating this expression:
Heat = 3.80 mol × 554 kJ
Therefore, the heat absorbed when 3.80 mol of A reacts is 2109.2 kJ (rounded to one decimal place).