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xxTIMURxx [149]
3 years ago
15

A chamber with a fixed volume is shown above. The temperature of the gas inside the chamber before heating is 25.2 C and it’s pr

essure is 0.600 atm. The gas is heated with a flame to a temperature of 72.4 C, what is it’s pressure at this temperature
Chemistry
1 answer:
rusak2 [61]3 years ago
3 0

Answer:

Explanation:

Given parameters:

Initial temperature T₁  = 25.2°C  = 25.2 + 273  = 298.2K

Initial pressure  = P₁  = 0.6atm

Final temperature = 72.4°C   = 72.4 + 273  = 345.4K

Unknown:

Final pressure = ?

Solution:

To solve this problem, we use an adaption of the combined gas law where the volume gas is fixed. This simplification results into:

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

where P and T are temperatures, 1 and 2 are initial and final temperatures.

 Input the parameters and solve;

          \frac{0.6}{298.2}   = \frac{P_{2} }{345.4}  

          P₂   = 0.7atm

         

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Answer:

T = 215.33 °C

Explanation:

The activation energy is given by the Arrhenius equation:

k = Ae^{\frac{-Ea}{RT}}

<u>Where:</u>

k: is the rate constant

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We have for the uncatalyzed reaction:

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And for the catalyzed reaction:

Ea₂ = 42 kJ/mol

T₂ = 20 °C = 293 K

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Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

k_{1} = k_{2}

Ae^{\frac{-Ea_{1}}{RT_{1}}} = Ae^{\frac{-Ea_{2}}{RT_{2}}}   (1)

By solving equation (1) for T₁ we have:

T_{1} = \frac{T_{2}*Ea_{1}}{Ea_{2}} = \frac{293 K*70 kJ/mol}{42 kJ/mol} = 488. 33 K = 215.33 ^\circ C  

Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.

I hope it helps you!      

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