Question

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?

Answer 1

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) = ??

Answer 2

At last, someone comes up with the "right" anwesr!

Answer 3

Substance

(
-253
-417
195
-475

Answer 4

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

Answer 5

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

Answer 6

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.

Answer 7

[AI-generated answer — Explain Bot]

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.