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Nonamiya [84]
3 years ago
7

What celsius temperature, t2, is required to change the volume of the gas sample in part a (t1 = 43 ∘c , v1= 1.13×103 l ) to a v

olume of 2.26×103 l ? assume no change in pressure or the amount of gas in the balloon?
Chemistry
1 answer:
mamaluj [8]3 years ago
6 0
In this case, you are given the temperature and volume of the gas. To answer this question, you will need to use PV=nRT formula. Since the number of molecule and pressure is not changed then you can put 1 in both equations. The formula will become:

PV=nRT 
1 V= 1 T
V=T
T/V= 1

Dont forget that the celcius unit should be converted into kelvin. Then the calculation to find the temperature would be: 
T1/V1 = T2/V2
(43+273)/ 1.13x10^3 = T2/ 2.26 x 10^3
T2= (2.26 * 10^3) / (1.13 * 10^3) * (43+273)
T2= 2* 316= 632° K= 359°C
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Answer:

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As Q < Kp, the reaction didn't reach the equilibrium, and the value must increase. As we can notice by the equation, Q is directly proportional to the partial pressure of BrCl, so it must increase, and be greater than 2.00 atm in the equilibrium.

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Because is a reversible reaction, it will not go to completion, it will reach an equilibrium, and as discussed above, the partial pressures will change.

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