Enthalpy and Calorimetry: Quiz

Enthalpy and Calorimetry: Quiz

7 questions

1. Question 1EasyWhich process is endothermic?

   ABurning methaneBNeutralizing HCl with NaOHCDissolving ammonium nitrate in water (a cold pack)DRespiration
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   Answer: Dissolving ammonium nitrate in water (a cold pack)

   The cold pack absorbs heat from its surroundings, so it feels cold: ΔH is positive. The others release heat.

2. Question 2EasyFor an exothermic reaction, which is true?

   AΔH > 0 and the surroundings cool downBΔH < 0 and the surroundings warm upCΔH > 0 and the surroundings warm upDΔH < 0 and the surroundings cool down
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   Answer: ΔH < 0 and the surroundings warm up

   Heat flows out of the reaction into the surroundings, so the products are lower in enthalpy: ΔH is negative.

3. Question 3MediumHow much heat is needed to raise the temperature of 40.0 g of copper (c = 0.385 J/(g·°C)) from 25.0 °C to 75.0 °C?

   A770. JB15.4 JC1.16 × 10³ JD8.37 × 10³ J
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   Answer: 770. J

   q = (40.0 g)(0.385 J/(g·°C))(50.0 °C) = 770. J. The answer 8.37 × 10³ J uses the specific heat of water.

4. Question 4MediumIn a calorimeter, 100. g of solution warms by 6.8 °C during a reaction (c = 4.184 J/(g·°C)). What is q for the reaction?

   A+2.8 kJB−2.8 kJC+28 kJD−0.68 kJ
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   Answer: −2.8 kJ

   q(solution) = (100. g)(4.184 J/(g·°C))(6.8 °C) = 2845 J = +2.8 kJ, so q(reaction) = −2.8 kJ: the reaction released the heat.

5. Question 5EasyCH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = −890.5 kJ. What is ΔH for 2CH₄ + 4O₂ → 2CO₂ + 4H₂O?

   A−890.5 kJB−1781.0 kJC−445.3 kJD+1781.0 kJ
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   Answer: −1781.0 kJ

   Doubling the equation doubles ΔH: 2 × (−890.5 kJ) = −1781.0 kJ.

6. Question 6MediumWhy do simple coffee-cup calorimeters usually give ΔH values slightly smaller in size than accepted values?

   AWater has no specific heatBSome heat is lost to the surroundings and the cupsCThe reaction goes faster in polystyreneDTemperature is measured in °C instead of K
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   Answer: Some heat is lost to the surroundings and the cups

   Heat that escapes is not counted in q = mcΔT, so the measured heat change is a little too small.

7. Question 7Medium0.0500 mol of water forms in a neutralization that releases 2.85 kJ. What is ΔH per mole of water?

   A−0.143 kJ/molB−57.0 kJ/molC+57.0 kJ/molD−2.85 kJ/mol
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   Answer: −57.0 kJ/mol

   ΔH = −2.85 kJ ÷ 0.0500 mol = −57.0 kJ/mol. It is negative because heat was released.