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) = ??
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?
6 answers
At last, someone comes up with the "right" anwesr!
Substance
(
-253
-417
195
-475
(
-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
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
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
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
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.