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grandymaker [24]
4 years ago
13

Calculate the Equilibrium constant Kc at 25 C from the free - energy change for the following reaction . Zn(s) + 2Ag+(aq)<---

----> Zn2+(aq)+ Ag(s)
Chemistry
1 answer:
FinnZ [79.3K]4 years ago
5 0

Answer:  Kc=1.24*10^5^2

Explanation: For the given reaction:

Kc=\frac{[Zn^+^2]}{[Ag^+]^2}

Concentrations of the ions are not given so we need to think about another way to calculate Kc.

We can calculate the free energy change using the standard cell potential as:

\Delta G^0=-nFE^0_c_e_l_l

E^0_c_e_l_l can be calculated using standard reduction potentials.

Standard reduction potential for zinc is -0.76 V and for silver, it is +0.78 V.

E^0_c_e_l_l = E^0_c_a_t_h_o_d_e+E^0_a_n_o_d_e

Reduction takes place at anode and oxidation at cathode. As silver is reduced, it is cathode. Zinc is oxidized and so it is anode.

E^0_c_e_l_l  = 0.78 V - (-0.76 V)

E^0_c_e_l_l  = 0.78 V + 0.76 V

E^0_c_e_l_l  = 1.54 V

Value of n is two as two moles of electrons are transferred in the cell reaction F is Faraday constant and its value is 96485 C/mol of electron .

\Delta G^0=-(2*96485*1.54)

\Delta G^0 = -297173.8 J

Now we can calculate Kc using the formula:

\Delta G^0=-RTlnKc

T = 25+273 = 298 K

R = 8.314JK^-^1mol^-^1

--297173.8 = -(8.314*298)lnKc

297173.8 = 2477.572*lnKc

lnKc=\frac{297173.8}{2477.572}

lnKc = 119.946

Kc=e^1^1^9^.^9^4^6

Kc=1.24*10^5^2

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Answer: 234.4K

Explanation:

Given that,

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[Convert 20.0°C to Kelvin by adding 273

20.0°C + 273 = 293K]

New volume of gas (V2) = 4.0L

New temperature of gas (T2) = ?

Since volume and temperature are given while pressure is held constant, apply the formula for Charle's law

V1/T1 = V2/T2

5.00L/293K = 4.0L/T2

To get the value of T2, cross multiply

5.00L x T2 = 293K x 4.0L

5.00L•T2 = 1172L•K

Divide both sides by 5.00L

5.00L•T2/5.00L = 1172L•K/5.00L

T2 = 234.4K

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8 0
3 years ago
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A 1.50 g sample of solid NH₄NO₃ was added to 35.0 mL of water in a styrofoam cup (insulated from the environment) and stirred un
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Answer : The heat of the reaction is, 1.27 kJ/mole

Explanation :

First we have to calculate the heat released.

Formula used :

Q=m\times c\times \Delta T

or,

Q=m\times c\times (T_2-T_1)

where,

Q = heat = ?

m = mass of sample = 1.50 g

c = specific heat of water = 4.81J/g^oC

T_1 = initial temperature  = 22.7^oC

T_2 = final temperature  = 19.4^oC

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

Q=1.50g\times 4.81J/g^oC\times (19.4-22.7)^oC

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Therefore, the heat of the reaction is, 1.27 kJ/mole

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b

.................,

......

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