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xxMikexx [17]
3 years ago
15

A 2.00-mol sample of hydrogen gas is heated at constant pressure from 294 K to 414 K. (a) Calculate the energy transferred to th

e gas by heat. kJ (b) Calculate the increase in its internal energy. kJ (c) Calculate the work done on the gas. kJ
Chemistry
1 answer:
Furkat [3]3 years ago
6 0

Answer:

a) The energy transferred is 6.91 kJ

b) The internal energy is 4.90 kJ

c) The work done on the gas is - 2.01 kJ

Explanation:

Step 1: Data given

Number of moles of hydrogen gas = 2.00 moles

Pressure = constant

Temperature is heated from 294 K to 414 K

Molar heat capacity of hydrogen gas = 28.8 J/mol*K

Step 2: Calculate the energy transferred to the gas by heat.

Q = n* Cp * ΔT

⇒with Q =the energy transferred

⇒with n = the number of moles = 2.00 moles

⇒with Cp = the Molar heat capacity of hydrogen gas = 28.8 J/mol*K

⇒ with ΔT = Temperature 2 - Temperature 1 = 414 - 294 = 120K

Q = 2.00 * 28.8 * 120

Q = 6912 J = 6.91 kJ

Step 3: Calculate the increase in its internal energy.

ΔEint = n*Cv*ΔT

⇒with ΔEint = the increase in its internal energy.

⇒with n = the number of moles = 2.00 moles

⇒with Cv = The constant volume = 20.4 J/mol*K

⇒with  ΔT = Temperature 2 - Temperature 1 = 414 - 294 = 120K

ΔEint = 2.00 * 20.4 * 120

ΔEint =4896 J = 4.90 kJ

Step 4: Calculate the work done on the gas.

Work done on the gas = -Q + ΔEint

W = -6.91 kJ + 4.90 kJ

W = -2.01 kJ

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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:

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For some reversible reactions that involve gases:

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Explanation:

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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.

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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.

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Answer:

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