Heat of dilution of sulfuric acid
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What is the heat of dilution of sulfuric acid? How much heat needs to be released to dilute 1 ton of 98.5% concentrated sulfuric acid to 74% concentration? It is best to provide a detailed calculation process.
This post was last edited by dreamclear on 2009-5-14 at 10:57. This shouldn’t be very simple; a lot of the heat released in reality needs to be estimated. However, it should be easy to calculate based on theory. This calculation can likely be found in the October 1963 issue of \"Sulfuric Acid Industry\"
This post was last edited by dreamclear on 2009-5-14 at 10:56, with specific calculation procedures and steps provided for the original poster, in the hope that they would be able to learn from them. Enthalpy of dilution: The heat effect that occurs when a certain amount of solvent is added to a solution of a definite concentration is called the enthalpy of dilution. The enthalpy of dilution released when the concentration of sulfuric acid is reduced from C1 to C2 is given by Q2-Q1 = n2*17860/(n2+1.7983) – n1*17860/(n2+1.7983). Here, Q2 and Q1 represent the enthalpy of dilution per mole of sulfuric acid, in KJ/mol; n2 and n1 represent the number of moles of water per mole of sulfuric acid at concentrations C2 and C1 respectively. From this, it can be seen that when the concentration of sulfuric acid is 98.5%, the number of moles of water is 1.5/18*98/98.5 = 0.0829. When the concentration is 74%, the number of moles of water is 26/18*98/74 = 1.9129. Substituting these values into the formula gives Q2-Q1 = 8418.7023 KJ/mol. Therefore, based on the parameters provided by the original poster, the total heat released is 8418.7023 KJ/mol
Q2-Q1=n2*17860/(n2+1.7983) - n1*17860/(n2+1.7983) Should be: Q2-Q1=n2*17860/(n2+1.7983) - n1*17860/(n1+1.7983)
Reply to 3# dreamclear: You have only calculated the heat released per mole of dilute sulfuric acid, but not the total amount of heat released! The total heat released, Q = 1000000/98*8418.7 KJ/mol
Reply to 5# Jinxiangren: The \"Design Manual for Ammonium Phosphate Calcium Carbonate\" on P249 also provides the specific calculation process, which is exactly the same as that in the original post
Hehe, the \"Sulfuric Acid Work Manual\" also contains detailed calculation methods!
Reply to 6# 01wang*an: Are you working with sulfuric acid?
Reply to 1# qingqingdehaife: Is the original poster dealing with the selection of sulfuric acid diluters? We have just built one as well!
This post was last edited by 01wang*an on 2012-2-22 at 11:04. Page P249 of the “Design Manual for Ammonium Phosphate Calcium Carbonate” also provides the specific calculation process, which is different from what the original poster used. The formula is: Q = 74.76N2/(N2 + 1.7983) – 74.76*N1/(N1 + 1.7983), where Q represents the heat of dilution per mole of sulfuric acid, in KJ/mol; N1 is the molar ratio of water to sulfuric acid before dilution, and N2 is the molar ratio after dilution. The original poster’s formula was Q2 – Q1 = n2*17860/(n2 + 1.7983) – n1*17860/(n1 + 1.7983). The coefficients are different; let’s see where the issue lies
The unit is wrong; it should be joules and calories