1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
yarga [219]
1 year ago
5

Why is there a discrepancy between a heat of reaction ob-tained from calorimetry and one obtained from bond energies?

Chemistry
1 answer:
Eva8 [605]1 year ago
7 0

While bond energies and bond enthalpies can be used to estimate the heat of reaction (enthalpy change of a reaction), H, the heat of neutralization is the heat released when 1 mole of water is generated by the reaction of an acid and a base (reaction).

For the same type of bond, bond enthalpies differ from compound to compound. For instance, the C-H bond enthalpy in methane is nearly identical to that of ethane, butane, etc. When we look up the bond enthalpy for a C-H bond in a table of bond enthalpies, the average number that results may only be accurate to two or three significant figures.

Each compound's enthalpies of production are listed, and those numbers take into account any minor variations in the enthalpies of each bond. Therefore, the result will be more accurate if you utilize formation enthalpies rather than average bond enthalpies to compute a given reaction's enthalpy change.

Learn more about Bond enthalpy here-

brainly.com/question/9998007

#SPJ4

You might be interested in
SCIENTIFIC METHOD
hammer [34]
Smhsmz bomb b V mzvznshx
4 0
3 years ago
Read 2 more answers
ΔH for the reaction below is -826.0 kJ/mol. Calculate the heat change when a 69.03-g sample of iron is reacted.4Fe(s) + 3 O2(g)
Anuta_ua [19.1K]

Answer:

c. -1020.9 kJ

Explanation:

4Fe (s) + 3 O₂ (g) --> 2 Fe₂O₃(s)         ΔH  =  -826.0 kJ/mol.

atomic weight of iron = 56

69.03 g = 69.03 / 56

= 1.23268 moles

Heat released by 1.23268 moles

= 1.23268 x 826.0

= -1020.9 kJ .

4 0
3 years ago
Convert the pressure 525.4 torr to kPa.
Oduvanchick [21]

Answer:

Option A. 70.0 KPa.

Explanation:

Data obtained from the question include:

Pressure (torr) = 525.4 torr

Pressure (kPa) =?

The pressure expressed in torr can be converted kPa as shown below:

760 torr = 101.325 KPa

Therefore,

525.4 torr = (525.4 x 101.325) / 760 = 70.0 KPa.

Therefore, 525.4 torr is equivalent to 70.0 KPa.

3 0
3 years ago
Read 2 more answers
What is the percent of Cr in Cr2O3
IgorLugansk [536]
The percentage of Chromium in Chromium Oxide is calculated as follow,

Step 1:
           Calculate Molar mass of Cr₂O₃,
Cr = 51.99 u
O = 16 u
So,
     2(51.99) + 3(16) = 103.98 + 48 = 151.98 u

Step 2:
          Secondly divide molar mass of only chromium with total mass of Cr₂O₃ and multiply with 100.
i.e.
                   = \frac{103.98}{151.98} × 100
                   
                   = 68.41 %
So, the %age composition of chromium in chromium oxide is 68.41 %.
5 0
3 years ago
Read 2 more answers
I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol
galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = \Delta H_f_{(I_2)}=62.438 kJ/mol

Enthalpy of formation of chlorine gas = \Delta H_f_{(Cl_2)}=0 kJ/mol

Enthalpy of formation of ICl gas = \Delta H_f_{(ICl)}=17.78 kJ/mol

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]

=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol

Enthaply change when 1.62 moles of iodine gas recast:

\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ

Entropy of the surrounding = \Delta S^o_{surr}=\frac{\Delta H}{T}

=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

4 0
3 years ago
Other questions:
  • By mistake, you added salt instead of sugar to the oil. How can you remove the salt?
    7·2 answers
  • What happens when an atom gains an electron
    15·2 answers
  • Which of the following statements best describes the energy transformation that occurs when a log burns? Mechanical energy chang
    14·2 answers
  • Cobalt-60 is a strong gamma emitter that has a half- life of 5.26 yr. The co balt-60 in a radiotherapy unit must be replaced whe
    6·1 answer
  • Select the correct answer.
    13·2 answers
  • Chlorine + lithium iodide<br> →
    14·1 answer
  • 6.
    8·1 answer
  • I WILL GIVE A LOT OF EXTRA POINTS. PLEASE ANSWER ALL OF THEM ​
    7·2 answers
  • The surfaces of leaves contain guard cells that open and close the stomata.
    13·2 answers
  • How many H2O molecules are there in 4.5 g of water (see picture)
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!