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Answer:
The value of at 4224 K is 314.23.
Explanation:
Initially
4.97 atm 0
At equilibrium
4.97 - p 2p
At initial stage, the partial pressure of oxygen gas = =4.97 atm
At equilibrium, the partial pressure of oxygen gas =
So, 4.97 - p = 0.28 atm
p = 4.69 atm
At equilibrium, the partial pressure of O gas =
The expression of is given as :
The value of at 4224 K is 314.23.
Answer:
Temperature required = 923K
Explanation:
The question is incomplete as there are some details that has to be given. details like the values of the standard enthalpies and entropies of the reactants and product as this is needed to calculate the actual value of the standard enthalpies and standard entropies of the reaction. I was able to get those values from literature and then calculated what needs to be calculated.
From there, I was able to use the equation that shows the relationship between, gibb's free energy, enthalpy, entropy and temperature. The necessary mathematical manipulation were done and the values were plugged in to get the temperature required to make the reaction spontaneous.
A few notes on the Gibb's free energy.
The Gibb's free energy also referred to as the gibb's function represented with letter G. it is the amount of useful work obtained from a system at constant temperature and pressure. The standard gibb's free energy on the other hand is a state function represented as Delta-G, as it depends on the initial and final states of the system.
The spontaneity of a reaction is explained by the standard gibb's free energy.
- If Delta-G = -ve ( the reaction is spontaneous)
- if Delta -G = +ve ( the reaction is non-spontaneous)
- if Delta-G = 0 ( the reaction is at equilibrium)
The step by step calculations is done as shown in the attachment.
