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?
So for part on you will use the Delta H of products subtracted by the Delta H of reactants. using the values given and multiplying by the coefficient of the compound.
A. [2(203)+(-523)]-[2(-263)+(-391)] = Delta Hrxn which would be 932kJ
for part B the difference of the two moles from the first problem then the new moles giving then will be multiplied by the answer from part A
B. 3.8mols/2moles = 1.9 moles 1.9moles x 932kJ = H H = 1770.8kJ
At last, someone comes up with the "right" anwesr!
Substance
(
-253
-417
195
-475
Calculate the standard enthalpy change for the reaction
2A+B⇌2C+2D
Use the following data:
Substance ΔH∘f
(kJ/mol)
A -275
B -389
C 195
D -503
Express your answer to three significant figures and include the appropriate units.
Hints
Calculate how much energy in kJ is released per mole. 217 kJ is released for every 2 moles of A, determine how many moles are released per one mole of A. Should be 108.5 kJ. Simply multiply that by 3.80 moles to get 412.3 kJ released.
To calculate the heat absorbed when 3.80 mol of A reacts, we need to use the stoichiometry of the reaction and the enthalpy change.
Given:
A -263 kJ/mol
From the balanced equation:
2A + B -> 2C + 2D
We can see that for every 2 moles of A that react, 2C and 2D are produced. Therefore, the stoichiometric ratio is 1:1 between A and C (as well as between A and D).
To calculate the heat absorbed, we need to find the number of moles of C (nC) produced when 3.80 mol of A reacts. Since the stoichiometric ratio between A and C is 1:1, we can say that nC = 3.80 mol.
Since the heat change is given per mole of substance, we can use the enthalpy change (Delta H) for C, which is +203 kJ/mol.
Now we can calculate the heat absorbed using the formula:
Heat absorbed = Delta H * nC
Heat absorbed = +203 kJ/mol * 3.80 mol
Heat absorbed = +771.4 kJ
Therefore, the heat absorbed when 3.80 mol of A reacts is 771.4 kJ.
I have an answer of 271 kJ/rxn for part A. Check your math.
For part B,
271 kJ/2 mol x (3.80 mol) = ??