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guapka [62]
2 years ago
6

6. From the values of ΔH and ΔS, predict which of the following reactions would be spontaneous at 25ºC: Reaction A: ΔH = 10.5 kJ

/mol, ΔS = 30 J/K ∙ mol Reaction B: ΔH = 1.8 kJ/mol, ΔS = –113 J/K ∙ mol If any of the above reactions is nonspontaneous at 25ºC, at what temperature might it become spontaneous? (16 points)
Chemistry
1 answer:
qwelly [4]2 years ago
3 0

Answer:

Both reaction A and reaction B are non spontaneous.

Explanation:

For a spontaneous reaction, change in gibbs free energy (\Delta G) should be negative.

We know, \Delta G=\Delta H-T\Delta S, where T is temperature in Kelvin scale.

Reaction A: \Delta G=(10.5\times 10^{3})-(298\times 30)J/mol=1560J/mol

As \Delta G is positive therefore the reaction is non-spontaneous.

If at a temperature T K , the reaction is spontaneous then-

\Delta H-T\Delta S< 0

or, T> \frac{\Delta H}{\Delta S}

or, T> \frac{10.5\times 10^{3}}{30}

or, T> 350

So at a temperature greater than 350 K, the reaction is spontaneous.

Reaction B: \Delta G=(1.8\times 10^{3})-(-113\times 298)J/mol=35474J/mol

As \Delta G is positive therefore the reaction is non-spontaneous.

If at a temperature T K , the reaction is spontaneous then-

\Delta H-T\Delta S< 0

or, T> \frac{\Delta H}{\Delta S}

or, T> \frac{1.8\times 10^{3}}{-113}

or, T> -16

So at a temperature greater than -16 K, the reaction is spontaneous.

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40.3∘C

Explanation:

At planet B;

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Water freezes = 50∘C

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A hypothetical AX type of ceramic material is known to have a density of 3.55 g/cm3 and a unit cell of cubic symmetry with a cel
postnew [5]

Explanation:

For AX type ceramic material, the number of formula per unit cells is as follows.

         \rho = \frac{n'(A_{c} + A_{A})}{V_{C}N_{A}}

or,     n' = \frac{\rho a^{3}N_{A}}{(A_{c} + A_{A})}

where,   n' = no. of formula units per cell

           A_{c} = molecular weight of cation = 90.5 g/mol

           A_{a} = molecular weight of anion = 37.3 g/mol

          V_{c} = volume of cubic cell = 3.55 g/cm^{3}

           a = edge length of unit cell = 3.9 \times 10^{-8} cm

        N_{A} = Avogadro's number = 6.023 \times 10^{23}

          \rho = density = 3.55 g/cm^{3}

Now, putting the given values into the above formula as follows.

           n' = \frac{\rho a^{3}N_{A}}{(A_{c} + A_{A})}

                      = \frac{3.55 g/cm^{3} \times (3.9 \times 10^{-8})^{3} \times 6.023 \times 10^{23}}{(90.5 + 37.3)}

                     = 0.9

                    = 1 (approx)

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3 0
2 years ago
What is the pH of a 0.28 M solution of ascorbic acid (Vitamin C)? (The values for Ka1 and Ka2 for ascorbic acid are 8.0×10−5 and
Tomtit [17]

Answer:

pH = 2.32

Explanation:

H2A + H2O -------> H3O+ + HA-    

Ka2 is very less so i am not considering that dissociation.

now Ka = 8.0×10−5

            = [H3O+] [HA-] / [H2A]

lets concentration of H3O+ = X then above equation will be

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X2 + 8.0×10−5 X - 2.24 x 10−5

solve the quardratic equation

X =0.004693 M

pH = -log[H+}

    = -log [0.004693]

    = 2.3285

    ≅2.32

pH = 2.32

5 0
2 years ago
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