Answer : The value of for this reaction is,
Explanation :
The given chemical reaction is:
Now we have to calculate value of .
where,
= Gibbs free energy of reaction = ?
n = number of moles
= -389.8 kJ/mol
= -161.96 kJ/mol
= -137.2 kJ/mol
Now put all the given values in this expression, we get:
The relation between the equilibrium constant and standard Gibbs, free energy is:
where,
= standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol
R = gas constant = 8.314 J/L.atm
T = temperature =
= equilibrium constant = ?
Now put all the given values in this expression, we get:
Thus, the value of for this reaction is,
Here is the full question:
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.
What is the concentration at 10 minutes? (Round your answer to three decimal places.
Answer:
0.046 %
Explanation:
The rate-in;
= 0.8
The rate-out
=
=
We can say that:
where;
A(0)= 0.2% × 6000
A(0)= 0.002 × 6000
A(0)= 12
Integration of the above linear equation =
so we have:
∴
Since A(0) = 12
Then;
Hence;
∴ the concentration at 10 minutes is ;
= %
= 0.0456667 %
= 0.046% to three decimal places
Answer : The enthalpy of the reaction is, -2552 kJ/mole
Explanation :
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The given enthalpy of reaction is,
The intermediate balanced chemical reactions are:
(1)
(2)
(3)
(4)
Now we have to revere the reactions 1 and multiple by 2, revere the reactions 3, 4 and multiple by 2 and multiply the reaction 2 by 2 and then adding all the equations, we get :
(when we are reversing the reaction then the sign of the enthalpy change will be change.)
The expression for enthalpy of the reaction will be,
Therefore, the enthalpy of the reaction is, -2552 kJ/mole