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For the reaction given in Part A, How much heat is absorbed when 3.00mol of A reacts?

Express your answer numerically in kilojoules.

Part A (info):

Calculate the standard enthalpy change for the reaction

2A+B <=> 2C +2D

where the heats of formation are given in the following table:

Substance Delta H(f)

(kJ/mol)

A -247

B -399

C 219

D -507

Express your answer in kilojoules.

Answer: 317

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Total energy of formation for right side = 2 * -247 + -399 = -893

Total energy of formation for left side = 2 * 219 + 2 * -507 = -576

dH = -576-(-893) = 317 kJ/mol

This is for 1 mole of B or 2 mole of A per reaction. for 3 mole we have:

317 kJ/2molA * 3molA = 475.5 kJ

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