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When a 2.50 L flask containing 1.00 atm of gas A is joined with a 0.500 L flask...
When a 2.50 L flask containing 1.00 atm of gas A is joined with a 0.500 L flask containing 0.750 atm of gas B, and the stopcock between the flasks is opened, and the gases are allowed to mix thoroughly, the following reaction proceeds at constant temperature. What is the resulting total pressure of gases joined flasks after the reaction proceeds to completion?
A(g) + 3B(g) --> AB3(g)
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We can use ideal gas law here assuming both gases act as ideal gasses.
First we should find the moles of each gas before mixing.
`PV = nRT`
`n = (PV)/(RT)`
Assume the temperature is T in Kelvins.
For gas A;
`n_A = (1xx2.5)/(0.08206T) = 30.466/T`
For gas B;
`n_B = (0.75xx0.5)/(0.08206T) = 4.569/T`
`A+3B rarr AB_3`
To complete the reaction we need 3times B than A. But as you can see above it doen't have that much of B in the flask to react with all the A.
So what will happen is all the B will be reacted.
Amount of B reacted `= 4.569/T`
Amount of A reacted `= 1/3xx(4.569/T) = 1.523/T`
So the rest A in flask `= 30.466/T-1.523/T = 28.477/T`
Amount of `AB_3` formed = reacted amount of A
Amount of `AB_3` formed `= 1.523/T`
So in the final mixture we have excess A and produced AB3.
Total moles in the mixture `= (1.523/T+28.477/T)`
Volume of flask `= (2.5+0.5) = 3L`
Using PV = nRT
`P = (nRT)/V`
`P = (1.523/T+28.477/T)xx0.08206xxT/3 = 0.8206atm`
So the total pressure of the system after joining the flasks is 0.8206atm.
Posted by jeew-m on August 10, 2013 at 6:19 PM (Answer #1)
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