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Example #5) 10mL of 3.0M HBr is mixed with 20mL of 2.0M KOH HBr + KOH → H₂O + KBr Substance H₂O HBr KBr KOH AHf in kJ/mol - 286 36.2 -392 - 480

Example #5) 10mL of 3.0M HBr is mixed with 20mL of 2.0M KOH HBr + KOH → H₂O + KBr Substance H₂O HBr KBr KOH AHf in kJ/mol - 286 36.2 -392 - 480

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Example #5) 10mL of 3.0M HBr is mixed with 20mL of 2.0M KOH HBr +KOH → H₂O + KBr AHreaction=AHproducts-AHreactants Substance H₂O HBr KBr KOH Here's the first reactant: 10mL of 3.0M HBr 10 * 3 = 0.03 mol H Br קסיטן (-286+-392)+(-36.27 +2 480) == -161.8 kJ 1000 Let's calculate the limiting reagent to determine how many mols of water will be formed: * - Let's do this for the second reactant, 20mL of 2.0M KOH 20 2 = 0.04 molk@H Calculate the amount of energy released: 286 36.2 0·03 * - 161.8=-4-854 k AHf in kJ/mol -392 480
Transcribed Image Text:Example #5) 10mL of 3.0M HBr is mixed with 20mL of 2.0M KOH HBr +KOH → H₂O + KBr AHreaction=AHproducts-AHreactants Substance H₂O HBr KBr KOH Here's the first reactant: 10mL of 3.0M HBr 10 * 3 = 0.03 mol H Br קסיטן (-286+-392)+(-36.27 +2 480) == -161.8 kJ 1000 Let's calculate the limiting reagent to determine how many mols of water will be formed: * - Let's do this for the second reactant, 20mL of 2.0M KOH 20 2 = 0.04 molk@H Calculate the amount of energy released: 286 36.2 0·03 * - 161.8=-4-854 k AHf in kJ/mol -392 480
Example #6) 250mL of 2.0M HBr is mixed with 300mL of 1.5M KOH HBr + NaOH > H2O + NaBr Substance H₂O HBr NaBr NaOH AHreaction=AHproducts-AHreactants (-286 +-362)-(-36.2 +-42607) = -18531 kJ Here's the first reactant: 250mL of 2.0M HBr 250 * 2 1000 * 2 = 0.5 mol H Br Let's calculate the limiting reagent to determine how many mols of water will be formed: 1000 Let's do this for the second reactant, 300mL of 1.5M KOH 300 2 = 0.6 mol koff * Calculate the amount of energy released: AHf in kJ/mol -286 36.2 -362-8 -426.7 0.5 * - 1851 = -192.55k
Transcribed Image Text:Example #6) 250mL of 2.0M HBr is mixed with 300mL of 1.5M KOH HBr + NaOH > H2O + NaBr Substance H₂O HBr NaBr NaOH AHreaction=AHproducts-AHreactants (-286 +-362)-(-36.2 +-42607) = -18531 kJ Here's the first reactant: 250mL of 2.0M HBr 250 * 2 1000 * 2 = 0.5 mol H Br Let's calculate the limiting reagent to determine how many mols of water will be formed: 1000 Let's do this for the second reactant, 300mL of 1.5M KOH 300 2 = 0.6 mol koff * Calculate the amount of energy released: AHf in kJ/mol -286 36.2 -362-8 -426.7 0.5 * - 1851 = -192.55k
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