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ivann1987 [24]
4 years ago
13

Which statement about spontaneous and nonspontaneous processes is correct? Spontaneous processes are favored by a decrease in H,

but nonspontaneous reactions are favored by an increase in H. Spontaneous processes are favored by a decrease in S, but nonspontaneous reactions are favored by an increase in S. An increase in S favors both spontaneous and nonspontaneous processes. A decrease in H favors both spontaneous and nonspontaneous processes.
Chemistry
2 answers:
Marina86 [1]4 years ago
7 0
I believe your answer would be the first one

hope this helps
Sav [38]4 years ago
5 0

Answer: Spontaneous processes are favored by a decrease in H, but nonspontaneous reactions are favored by an increase in H.

Explanation:

\Delta G=\Delta H-T\Delta S

\Delta G = gibbs free energy

\Delta H = change in enthalpy

\Delta S = change in entropy

For a reaction to be spontaneous, \Delta G= -ve. For this \Delta H  has to be negative i.e. there should be a decrease in enthalpy and \Delta S has to be positive i.e. there should be a increase in entropy.

\Delta G=(-ve)-T(+ve)=-ve-ve=-ve


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Since we are given the mass, specific heat, and change in temperature, we should use this formula for heat:

q=mc\Delta T

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m= 450.0 \ g \\c= 1.264 \ J/g \textdegree C\\\Delta T= 7.1 \ \textdegree C

Substitute the values into the formula.

q= (450.0 \ g)(1.264 \ J/g \textdegree C)(7.1 \ \textdegree C)

Multiply the first 2 values together. The grams will cancel out.

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Which of the following most likely happens when the volume of a gas increases? (4 points)
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(have a great day!)

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Answer :

(a) The change in internal energy of the gas is 22.86 kJ.

(b) The change in enthalpy of the gas is 34.29 kJ.

Explanation :

(a) The formula used for change in internal energy of the gas is:

\Delta U=nC_v\Delta T\\\\\Delta U=nC_v(T_2-T_1)

where,

\Delta U = change in internal energy = ?

n = number of moles of gas = 5 moles

C_v = heat capacity at constant volume = 2R

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 25^oC=273+25=298K

T_2 = final temperature = 300^oC=273+300=573K

Now put all the given values in the above formula, we get:

\Delta U=nC_v(T_2-T_1)

\Delta U=(5moles)\times (2R)\times (573-298)

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\Delta U=22863.5J=22.86kJ

The change in internal energy of the gas is 22.86 kJ.

(b) The formula used for change in enthalpy of the gas is:

\Delta H=nC_p\Delta T\\\\\Delta H=nC_p(T_2-T_1)

where,

\Delta H = change in enthalpy = ?

n = number of moles of gas = 5 moles

C_p = heat capacity at constant pressure = 3R

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 25^oC=273+25=298K

T_2 = final temperature = 300^oC=273+300=573K

Now put all the given values in the above formula, we get:

\Delta H=nC_p(T_2-T_1)

\Delta H=(5moles)\times (3R)\times (573-298)

\Delta H=(5moles)\times 3(8.314J/mole.K)\times (573-298)

\Delta H=34295.25J=34.29kJ

The change in enthalpy of the gas is 34.29 kJ.

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