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vivado [14]
3 years ago
15

What does an energy diagram illustrate? Identify what the "hill" on the energy curve and the positions of the reactants and prod

ucts represent.
Chemistry
2 answers:
Korvikt [17]3 years ago
6 0

Answer:

It shows the energy requirements needed in order for a reaction to occur

Explanation:

k0ka [10]3 years ago
5 0
An energy diagram of a chemical reactions illustrates the changes of energy as the chemical reactions advances.

At first, the energy in the diagram is the energy of the reactants.

As the reaction goes forward, the reactants start to react forming a transition compound, with a maximum energy level on the graph. This is the hill. So the hill represents the Activation Energy.

After that, the energy starts to decrease and at the end you have the energy of the products, which may be higher or lower than the initial energy of the reactants, depending upon whether the reaction is exothermic or endothermic.

For exothermic reactions the energy level of the products is lower than the energy level of the reactants, while for endothermic reactions the energy level of the products is higher than the energy level of the reactants.
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Which properties determines the element that an atom Is?
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Explanation:

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The big number is the amount of protons of the atom, which define each atom. The smaller number represents the atomic mass of the atom.

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When the heat of reaction (H=ve), the reaction is Endothermic reaction<br> True<br> False
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3 years ago
One mole of a monatomic ideal gas is subjected to the following sequence of steps: a. Starting at 300 K and 10 atm, the gas expa
Verdich [7]

Answer:

a) Q = 0; W = 0; ΔU = 0; ΔH = 0; ΔS = 0.09 atm.L/K

b) Q = 1250 J; W = 0; ΔU = 1250 J; ΔH = 1250 J; ΔS = -0.0235 atm.L/K

c) Q = 3653.545 J; W = - 3653.545 J; ΔU = 0; ΔH = 0; ΔS = - 3653.545 J

d) Q = - 2080 J; W = 830 J; ΔU = - 1250 J; ΔH = - 2080 J; ΔS = - 5.984 J/K

Explanation:

a) If there is a vacuum, the work is zero, as it is a free expansion, the volume increases, the pressure decreases, the temperature is constant and the internal energy is constant.

∴ n = 1 mole

∴ PV = RTn....ideal gas

∴ P1 = 10 atm

∴ R = 0.082 atm.L/K.mol

∴ T = 300 K = T2

∴ V2 = 3*V1

⇒ W = 0.....expands freely into vacuum

⇒ ΔU = Q = 0....first law

⇒ ΔS = -  nR Ln(P2/P1).....ideal gas

∴ V1*P1/T1 = V2*P2/T2

∴ T1 = T2 = 300 K

⇒ P2 = V1*P1 / V2 = V1*P1 / 3V1 = 10 atm/3 = 3.33 atm

⇒ ΔS = - (1mol)*(0.082 atm.L/K.mol) Ln ( 3.33/10)

⇒ ΔS = 0.09 atm.L/K

∴ ΔH = ΔU + (P2V2 - P1V1) = 0 + 0 = 0

b) heated reversibly at constant volume:

⇒ W = 0 ...at constant volume

∴ T2 = 400 K; T1 = 300 K

∴ V1 = V2

⇒ Q = ΔU = CvΔT....first law

∴ Cv = 12.5 J/K.mol.....monoatomic ideal gas

∴ ΔT = 400 - 300 = 100 K

⇒ Q = ΔU = 12.5 J/mol.K * 100K = 1250 J/mol * 1 mol = 1250 J

∴ ΔH = ΔU + PΔV = ΔU + 0 = 1250 J

∴ ΔS = - nR Ln (P2/P1)

∴ P2/T2 = P1/T1...constant volume

∴ P1 = 3.33 atm

⇒ P2 = P1*T2 / T1 = (3.33 atm)*(400K) / (300K) = 4.44 atm

⇒ ΔS = - (1mol)*(0.082atm.L/K.mol) Ln (4.44/3.33)

⇒ ΔS = - 0.0235 atm.L/K

c) reversibly expanded at constant temperature:

∴ T1 = T2 = 400K

∴ V2 = 3*V1

∴ ΔU = 0...constant temperature

⇒ Q = - W....fisrt law

∴ W = - ∫ PdV..... reversibly expansion

∴ P = nRT/V... ideal gas

⇒ W = - nRT ∫ dV/V

⇒ W = - nRT Ln (V2/V1)

⇒ W = - (1mol)*(8.314 J/K.mol) Ln (3)

⇒ W = - 9.134 J/K *400K = - 3653.545 J

⇒ Q = - W = 3653.545 J

⇒ ΔH = ΔU + P1V1 - P2V2 = 0 + nRT1 - nRT2 = 0 + 0 = 0

∴ ΔS = - nR Ln(P2/P1)

∴ P1 = 4.44 atm

⇒ P2 = V1*P1*T2/ V2*T1 = V1*(4.44atm)*(400K) / (3.V1)*(400K)

⇒ P2 = 4.44atm/3 = 1.48 atm

⇒ ΔS = - (1mol)*(8.314 J/mol.K) Ln (1.48/4.44)

⇒ ΔS = -9.134J/K * 400K = - 3653.545 J

d) reversibly cooled at constant pressure:

∴ T2 = 300 K;  T1 = 400 K

∴ P2 = P1

⇒ Q = ΔH = CpΔT

∴ Cp = 20.8 J/K.mol

∴ ΔT = 300 - 400 = - 100 K

⇒ Q = ΔH = 20.8 J/mol.K * ( -100K) = - 2080 J/mol * 1mol = - 2080 J

⇒ ΔU = nCvΔT = (1mol)*(12.5 J/mol.K)*( - 100K) = -1250 J

⇒ W = ΔU - Q = ΔU - ΔH = -1250 J - ( - 2080 J ) = 830 J

∴ ΔS = ∫ δQ/T = ∫ nCpdT/T

⇒ ΔS = nCp Ln (T2/T1)

⇒ ΔS = (1mol)*(20.8 J/mol.K) Ln (300/400) = - 5.984 J/K

7 0
3 years ago
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