If a student performs an endothermic reaction in a calorimeter, how does the calculated value of \(\Delta H\) differ from the actual value if the heat exchanged with the calorimeter is not taken into account?

Short Answer

Expert verified
When the heat exchanged with the calorimeter is not taken into account during the calculation of ΔH for an endothermic reaction, the calculated ΔH value will be less than the actual value. This is because the calorimeter absorbs some of the heat that was initially intended to be absorbed by the reaction, leading to a greater actual ΔH value.

Step by step solution

01

Understand endothermic reaction and calorimeters

An endothermic reaction is a chemical reaction that absorbs heat from its surroundings. During this reaction, energy is taken in by the system, which causes an increase in temperature. A calorimeter is a device used for measuring the heat absorbed or released during a chemical reaction. It helps in calculating ΔH for a given reaction.
02

Understand calculating ΔH using a calorimeter

In ideal circumstances, when a reaction is performed in a calorimeter, the energy absorbed or released is only associated with the chemicals involved in the reaction. To calculate ΔH, we use the formula: ΔH = q / n where q is the heat exchanged (absorbed or released) during the reaction and n is the amount of substance (in moles) involved in the reaction.
03

Analyze the effect of not considering heat exchanged with the calorimeter

When a reaction is performed in a calorimeter, some heat may be exchanged between the reaction and the calorimeter itself. By not taking this heat exchange into account, we are making the assumption that the calorimeter is a perfect insulator – meaning it does not absorb or lose any heat energy. In reality, no calorimeter is perfectly insulated, so the calculated ΔH will deviate from the actual value. In an endothermic reaction, since the reaction is absorbing heat from its surroundings, if the calorimeter also absorbs some of this heat, the actual value of ΔH will be greater than the calculated value. This is because the calorimeter has absorbed some of the heat that was initially supposed to be absorbed by the reaction.
04

Conclusion

If the heat exchanged with the calorimeter is not taken into account during the calculation of ΔH for an endothermic reaction, the calculated value will be less than the actual value. The actual value of ΔH in an endothermic reaction would be greater because some heat is absorbed by the calorimeter, in addition to the heat absorbed by the reaction itself.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Which has the greater kinetic energy, an object with a mass of 2.0 \(\mathrm{kg}\) and a velocity of 1.0 \(\mathrm{m} / \mathrm{s}\) or an object with a mass of 1.0 \(\mathrm{kg}\) and a velocity of 2.0 $\mathrm{m} / \mathrm{s}$ ?

The combustion of 0.1584 g benzoic acid increases the temperature of a bomb calorimeter by \(2.54^{\circ} \mathrm{C}\) . Calculate the heat capacity of this calorimeter. (The energy released by combustion of benzoic acid is 26.42 \(\mathrm{kJ} / \mathrm{g} .\) ) A \(0.2130-\mathrm{g}\) sample of vanillin \(\left(\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}_{3}\right)\) is then burned in the same calorimeter, and the temperature increases by $3.25^{\circ} \mathrm{C}$ . What is the energy of combustion per gram of vanillin? Per mole of vanillin?

A biology experiment requires the preparation of a water bath at $37.0^{\circ} \mathrm{C}$ (body temperature). The temperature of the cold tap water is \(22.0^{\circ} \mathrm{C},\) and the temperature of the hot tap water is \(55.0^{\circ} \mathrm{C} .\) If a student starts with 90.0 \(\mathrm{g}\) cold water, what mass of hot water must be added to reach $37.0^{\circ} \mathrm{C} ?$

The complete combustion of acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}(g),\) produces 1300 . kJ of energy per mole of acetylene consumed. How many grams of acetylene must be burned to produce enough heat to raise the temperature of 1.00 gal water by \(10.0^{\circ} \mathrm{C}\) if the process is 80.0\(\%\) efficient? Assume the density of water is 1.00 \(\mathrm{g} / \mathrm{cm}^{3}\)

A balloon filled with 39.1 moles of helium has a volume of 876 \(\mathrm{L}\) at \(0.0^{\circ} \mathrm{C}\) and 1.00 atm pressure. The temperature of the balloon is increased to \(38.0^{\circ} \mathrm{C}\) as it expands to a volume of 998 \(\mathrm{L}\) , the pressure remaining constant. Calculate \(q, w,\) and \(\Delta E\) for the helium in the balloon. (The molar heat capacity for helium gas is 20.8 \(\mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{mol.} )\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free