Is the flame test a test for the cation (metal listed first) or the anion (nonmetal listed second)?
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The question is incomplete, here is the complete question:
While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures.
A chemical engineer studying this reaction fills a 1.5 L flask at 12°C with 1.8 atm of ethylene gas and 4.7 atm of water vapor. When the mixture has come to equilibrium she determines that it contains 1.16 atm of ethylene gas and 4.06 atm of water vapor.
The engineer then adds another 1.2 atm of ethylene, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.
<u>Answer:</u> The partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm
<u>Explanation:</u>
We are given:
Initial partial pressure of ethylene gas = 1.8 atm
Initial partial pressure of water vapor = 4.7 atm
Equilibrium partial pressure of ethylene gas = 1.16 atm
Equilibrium partial pressure of water vapor = 4.06 atm
The chemical equation for the reaction of ethylene gas and water vapor follows:
<u>Initial:</u> 1.8 4.7
<u>At eqllm:</u> 1.8-x 4.7-x
Evaluating the value of 'x'
The expression of for above equation follows:
Putting values in above expression, we get:
When more ethylene is added, the equilibrium gets re-established.
Partial pressure of ethylene added = 1.2 atm
<u>Initial:</u> 2.36 4.06 0.64
<u>At eqllm:</u> 2.36-x 4.06-x 0.64+x
Putting value in the equilibrium constant expression, we get:
Neglecting the value of x = 13.41 because equilibrium partial pressure of ethylene and water vapor will become negative, which is not possible.
So, equilibrium partial pressure of ethanol = (0.64 + x) = (0.64 + 0.363) = 1.003 atm
Hence, the partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm