# Chemistry

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
MY ANSWER FOR PART A: Delta H =277 kJ

PART B----THIS ONE I NEED HELP ON
For the reaction given in Part A, how much heat is absorbed when 3.80 mol of A reacts?

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3. 👁 5,345
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) = ??

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

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3. Substance
(
-253
-417
195
-475

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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

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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

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