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AveGali [126]
2 years ago
13

A certain reaction is spontaneous at temperatures below 400. K but is not spontaneous at temperatures above 400. K. If ΔH° for t

he reaction is -20. kJ mol-1 and it is assumed that ΔH° and ΔS° do not change appreciably with temperature, then the value of ΔS° for the reaction is
Chemistry
1 answer:
jeka57 [31]2 years ago
7 0

Answer:

ΔS = 0.05 kJ /mol or 50 J /mol

Explanation:

As given that the reaction is spontaneous below 400 K and non spontaneous above 400K. Thus the reaction is at equilibrium at 400 K temperature.

At equilibrium the change in free energy ΔG is zero.

The relation between free energy change, enthalpy change, entropy change and temperature is

ΔG = ΔH - TΔS

Where

ΔH = enthalpy change

T = temperature

ΔS = entropy change

at equilibrium (400 K)

ΔG = 0 = (-20) + 400XΔS

ΔS = 0.05 kJ /mol or 50 J /mol

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3 years ago
1. Starting with 9.3 moles of O 2 , how many moles of H 2 S will be needed and how many moles of SO 2 will
enyata [817]

Answer:

Moles of H₂S needed = 6.2 mol

Moles of SO₂ produced = 6.2 mol

Explanation:

Given data:

Number of moles of O₂ = 9.3 mol

Moles of H₂S needed = ?

Moles of SO₂ produced = ?

Solution:

Chemical equation:

2H₂S + 3O₂      →    2SO₂ + 2H₂O

Now we will compare the moles of oxygen with H₂S.

                  O₂             :           H₂S

                    3             :             2

                   9.3            :         2/3×9.3 = 6.2 mol

Now we will compare the moles of SO₂ with both reactant.

                   O₂             :            SO₂

                    3              :               2

                   9.3            :         2/3×9.3 = 6.2 mol      

                 H₂S             :            SO₂

                    2              :               2

                   6.2            :          6.2 mol      

So 6.2 moles of  SO₂ are produced.

         

6 0
3 years ago
A gas balloon has a volume of 106.0 liters when the temperature is 25.0 °C and the pressure is 740.0 mm Hg. What will its volume
ipn [44]

Answer : The final volume of gas will be, 103.3 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 740.0 mmHg  = 98.4 kPa

Conversion used : (1 mmHg = 0.133 kPa)

P_2 = final pressure of gas = 99.3 kPa

V_1 = initial volume of gas = 106.0 L

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 25.0^oC=273+25.0=298K

T_2 = final temperature of gas = 20.0^oC=273+20.0=293K

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

\frac{98.4kPa\times 106.0L}{298K}=\frac{99.3kPa\times V_2}{293K}

V_2=103.3L

Therefore, the final volume of gas will be, 103.3 L

8 0
3 years ago
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