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wel
3 years ago
9

If hydrogen were used as a fuel, it could be burned according to the following reaction: H2(g)+1/2O2(g)→H2O(g). Use average bond

energies to calculate ΔHrxn for this reaction Use average bond energies to calculate ΔHrxn for the combustion of methane (CH4). Which fuel yields more energy per mole and per gram?
Chemistry
1 answer:
serious [3.7K]3 years ago
7 0
The deltaHrxn = -243 kJ/mol the deltaHrxn of CH4(methane) = -802 kJ/mol 

The fuel that yields more energy per mole is METHANE. The negative sign merely signifies the release of energy. Thus, 802 kJ/mol is greater than 243 kJ/mol.

The fuel that yields more energy per gram is HYDROGEN. Here is the computation:
deltaHrxn = (-243 kJ/mol)(1 mol/2.016 g H2)  <span>= -120.535714286 kJ/g or -121 kJ/g

</span>deltaHrxn of CH4(methane) = (-802 kJ/mol)(1 mol/16.04 g) 
<span>= -50 kJ/g 
</span>
As discussed the negative sign serves as the symbol of released energy. Thus, 121 is greater than 50.
You might be interested in
What factors affect the dynamic state of equilibrium in a chemical reaction and how?
yanalaym [24]

Answer:

Only changes in temperature will influence the equilibrium constant K_c. The system will shift in response to certain external shocks. At the new equilibrium Q will still be equal to K_c, but the final concentrations will be different.

The question is asking for sources of the shocks that will influence the value of Q. For most reversible reactions:

  • External changes in the relative concentration of the products and reactants.

For some reversible reactions that involve gases:

  • Changes in pressure due to volume changes.

Catalysts do not influence the value of Q. See explanation.

Explanation:

\displaystyle K_c = {e}^{\Delta G/(R\cdot T)}.

Similar to the rate constant, the equilibrium constant K_c depends only on:

  • \Delta G the standard Gibbs energy change of the reaction, and
  • T the absolute temperature (in degrees Kelvins.)

The reversible reaction is in a dynamic equilibrium when the rate of the forward reaction is equal to the rate of the backward reaction. Reactants are constantly converted to products; products are constantly converted back to reactants. However, at equilibrium Q = K_c the two processes balance each other. The concentration of each species will stay the same.

Factors that alter the rate of one reaction more than the other will disrupt the equilibrium. These factors shall change the rate of successful collisions and hence the reaction rate.

  • Changes in concentration influence the number of particles per unit space.
  • Changes in temperature influence both the rate of collision and the percentage of particles with sufficient energy of reaction.

For reactions that involve gases,

  • Changing the volume of the container will change the concentration of gases and change the reaction rate.

However, there are cases where the number of gases particles on the reactant side and the product side are equal. Rates of the forward and backward reaction will change by the same extent. In such cases, there will not be a change in the final concentrations. Similarly, catalysts change the two rates by the same extent and will not change the final concentrations. Adding noble gases will also change the pressure. However, concentrations stay the same and the equilibrium position will not change.

8 0
3 years ago
I need help please ​
Alex17521 [72]

Answer: If you think about it, B. would be the most reasonable answer with the given factors.

4 0
3 years ago
Lyle filled a glass bottle completely with water. He put the lid on tightly, then put the water in the freezer to cool it down q
svlad2 [7]

Answer:

The answer would be  It breaks them up.

Explanation:

7 0
2 years ago
Given that the nucleophilic substitution reaction used 5.0 mL of t-pentyl alcohol and 12.0 mL of conc. hydrochloric acid to prod
Vladimir [108]

Answer:

4.90 g

Explanation:

Given that:

volume of t-pentyl alcohol = 5 mL

the standard density of t-pentyl alcohol = 0.805 g/mL

Recall that:

density = mass(in wt) /volume

mass = density × volume

mass = 0.805 g/mL × 5 mL

mass = 4.03 g

Volume of HCl used = 12 mL

The reaction for this equation is shown in the image attached below.

From the reaction,

88.15 g of t-pentyl alcohol reacts with concentrated HCl to yield 106.59 g pf t-pentyl chloride.

4.03 g of t-pentyl alcohol forms,

= \dfrac{106.59 \ g \times 4.03 \ g}{88.15 \ g} of t-pentyl chloride.

Therefore,

Theoretical yield of t-pentyl chloride = 4.90 g

8 0
3 years ago
YO I NEED HELP!!!! APEX PLEASE
podryga [215]
B is the correct answer for it
8 0
3 years ago
Read 2 more answers
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