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boyakko [2]
4 years ago
7

Need help !!!!! Stuck asap

Chemistry
1 answer:
Step2247 [10]4 years ago
4 0
<h2>Hello!</h2>

The answer is:

Hence, the new pressure will be 2.07 atm.

P_{2}=2.07atm

<h2>Why?</h2>

Since we know that the gas is inside of a rigid container, meaning that the volume will be kept constant, we can solve the problem using the Gay-Lussac's Law.

The the Gay-Lussac's Law establishes that when an ideal gas is kept at the same volume, the pressure and the temperature will be proportional.

We need to pay special attention when we are working with the Gay-Lussac's Law since its equaitons works with absolute temperatures (Kelvin ), so, if we are working with relative temperatures such as Celsius degrees or Fahrenheit degrees, we need to convert the temperatures to Kelvin.

We can convert from Celsius degrees to Kelvin using the following formula:

Temperature(K)=Temperature(C\°)+273

So, we have the Gay-Lussac's equation:

\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}

Also, we are given the following information:

T_{1}=30\° \\P_{1}=2atm\\T_{2}=40\°

Therefore, converting the temperature to Kelvin, we have:

T_{1}=30C\°=30+273K=303K\\\\T_{1}=40C\°=40+273K=313K\\

Now, calculating we have:

\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}

P_{2}=\frac{P_{1}}{T_{1}}*T_{2}\\\\P_{2}=\frac{2atm}{303K}*313K=2.07atm

Hence, the new pressure will be 2.07 atm.

P_{2}=2.07atm

Have a nice day!

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From the following reaction and data, find (a) S o of SOCl2 (b) T at which the reaction becomes nonspontaneous SO3(g) + SCl2(l)
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Answer:

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Explanation:

The  balanced reaction of interest is:

                           SO₃ (g) + SCl₂ (l) ⇒    SOCl₂ (l) +     SO₂ (g)

with the data:

ΔHºf (kJ/mol)      -396          -50.0          -245.6         -296.8

Sº(J/mol·K)             256.7       184               ?               -248.1

ΔGº=  -75.2 kJ

We know, we can find the standard  change inGibb´s free energy from the equation:

ΔGºrxn =  ΔHºrxn - TΔSºrxn

So we can calculate ΔHºrxn  = ∑ ΔHºf prod  -  ΔHºreact, and substitute into this equation to solve Sº SOCl₂.

ΔHºrxn = ( -245.6 + (-296.8) ) - ( -396 - 50) kJ = - 96.4 kJ

Similarly  for ΔSºrxn

 ΔSºrxn = (-0.248.1 +Sº SOCl₂) - (0.256.7 +0.184) kJ/K

= -0.689 kJ /K -+ Sº SOCl₂

Plugging the values for the expression for  ΔGºrxn:

-75.2 kJ = -96.4 kJ - 298 K  x  ( -0.689 kJ /K + Sº SCl₂ )

-75.2 kJ = -96.4 kJ + 205.3 Kj - 298 Sº SCl₂

-184  kJ = -298 K  x Sº SCl₂

0.618 kJ/molK = Sº SCl₂

= 0.618 kJ/K x 1000 J = 618 J/Kmol

For the second part we will still be using the Gibb´s free energy change  equation as above , but now we will solve for T when the reaction becomes  non-spontaneous.

For the reaction to become non-spontaneous  ΔGº is positive, and this happens when the term  TΔSº becomes greater tha ΔHº:

ΔGºrxn =  ΔHºrxn - TΔSºrxn

0 =   ΔHºrxn - TΔSºrxn ⇒  TΔSºrxn  =  ΔHºrxn

                                           T= ΔHºrxn / ΔSºrxn

ΔSºrxn  = -0.689 J/Kmol + 0.618 J/Kmol = -0.0710 kJ/Kmol

( using the value  the value just calculated from above )

T =  - 96.4 kJ / -0.071 kJ/K = 1.36 x 10³ K

For temperatures greater than 1.36 x 10³ K the reaction becomes non-spontaneous.

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3 years ago
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