Answer:
K2 +Br ->2KBr
K + I ->KI
actually I don't know the e option but I had tried can u pls balance it urself
By increasing atomic number
Answer:
Work done on the system is zero , hence no work is done since the process is <u>isochoric.</u> There is no work done if the volume remains unchanged. (Though the temperature rises, work is only accomplished when the volume of the gas changes.)
Explanation:
ISOCHORIC PROCESS - An isochoric process, also known as a constant-volume process, isovolumetric process, or isometric process, is a thermodynamic process in which the volume of the closed system undergoing the process remains constant through the process. The heating or cooling of the contents of a sealed, inelastic container is an example of an isochoric process. The thermodynamic process is the addition or removal of heat, the closed system is established by the isolation of the contents of the container, and the constant-volume condition is imposed by the container's inability to deform. It should be a quasi-static isochoric process in this case.
<u>Hence , the work done in the system is zero.</u>
Explanation:
Since, the given reaction is as follows.

Initial: 36.1 atm 0 0
Change: 2x x x
Equilibrium: (36.1 - 2x) x x
Now, expression for
of this reaction is as follows.
![K_{p} = \frac{[N_{2}][O_{2}]}{[NO]^{2}}](https://tex.z-dn.net/?f=K_%7Bp%7D%20%3D%20%5Cfrac%7B%5BN_%7B2%7D%5D%5BO_%7B2%7D%5D%7D%7B%5BNO%5D%5E%7B2%7D%7D)
As the initial pressure of NO is 36.1 atm. Hence, partial pressure of
at equilibrium will be calculated as follows.
![K_{p} = \frac{[N_{2}][O_{2}]}{[NO]^{2}}](https://tex.z-dn.net/?f=K_%7Bp%7D%20%3D%20%5Cfrac%7B%5BN_%7B2%7D%5D%5BO_%7B2%7D%5D%7D%7B%5BNO%5D%5E%7B2%7D%7D)
x = 18.1 atm
Thus, we can conclude that partial pressure of
at equilibrium is 18.1 atm.
<span>Who can help me with a few chemistry balancing equations?
Me. Just show me the equations :)</span>