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In this problem we are going to calculate the adiabatic flame temperature when normal
butane gas is burned in pure oxygen. We are going to feed both the oxygen and butane at
25 degrees C. Reaction is at 1 bar. Now the way we do this calculation is we create a
pathway so we can take an advantage of the idea of state function to make the calculation
easier. So the green path is drawn to represent that we know the exact temperature vs. amount
of conversion in the flame but we do know for the overall process since it is adiabatic
delta H is equal to 0. What we are going to do is pick a pathway where we start with butane
and oxygen at 298 we carry out the reaction. So we have heat of reaction at 298. We then
take the products which are CO2 and water and we raise their temperature to our final
temperature such that delta H change 1 plus delta H change 2 adds up to 0. This term is
going to be some summation. The number of moles times the integral of the heat capacity
from 298 to the final temperature. There are two terms here and I have written the general.
The number of moles of CO2 times it's heat capacity and number of moles of water times
it's heat capacity. So Cp is a function of temperature. This is important because we
are going to have a very high final temperature we have to account for the fact that the heat
capacity changes with temperature. We need to first the balanced equation. This balanced
equation means we have 1 moles of butane, 13/2 moles of oxygen, 4 moles of CO2, 5 moles
of water and of course for this final temperature here we have 4 moles of CO2 and 5 moles of
water.So first we are going to calculate delta H for the reaction. Delta H for a reaction,
this is at 298 Kelvin. It is going to be the delta H of formation of water times 5 plus
the delta H of formation of CO2 times 4 minus the delta H of formation of butane minus delta
H of formation of oxygen. Now this is zero because it's an element. So we can calculate
the heat of reaction. So substituting in the values from the tables for the heat of formation
at 298K I get the heat of reaction. This is in kilo Joules. Then we need the heat capacity
terms for CO2 and water. So we will look those up. So here our heat capacity is as a function
of temperature from a table where these are in Joules per mole per Kelvin. The temperature
we substitute in here is in Kelvin and what we need to do is multiply this by 4 and this
by 5. This is the corresponding number moles of CO2 and water in that product that's formed
and add these terms together. Number of moles times the heat capacity is equal to. So here
are the terms added together. Now we want to integrate this summation from our starting
temperature to our final temperature which is our adiabatic temperature. I have shown
the integration here and we can substitute in the two limits. So these are each now evaluated
from 298 and the final temperature. So the numbers were substituted in and then the equation
was simplified to a polynomial. Now we have almost finished. We need to go back and look
at our starting. We are going to add this term plus this term and it should add to 0
because state function. We picked a different pathway to get between the same initial and
final states. So now we are going to substitute in to that final relationship. So now all
we have to do is simplify this equation by combining these two terms and solving the
polynomial and we can do that in excel solver. We get a temperature of 3,867 Kelvin. Of course
it is not accurate to this many significant figures so around 3,600 degrees C. However
it is not clear that the heat capacities are very accurate at these higher temperature
and indeed the values reported for the adiabatic temperature for butane burned in pure oxygen
are significantly lower. More like 2,800 degrees C. So this is the adiabatic flame temperature
that calculate from heat of reaction and heat capacities that are a function of temperature.