8 replies to this topic

[shamir]

HOW TO CALCULATE HEAT OF REACTION FOR BELOW REACTION

 

 

P2O5( solid) + 3 H2O → 2 H3PO4(liquid)

(formation)        (formation)     ( formation)

 -360                  -68.32               (-306.2  ( all in kcal/gmole)

 

heat of reaction from above value  comes  = -42.4 cal/gmole.

 

During reaction p205 vapour temperature is 591 degree celcius and has to be cool down up to 60 degree celcius.

 

Now how to calculate heat required for condensation of P205 from 591 to 60 degree celcuis.

 

please help for calculating above problem.............

[PingPong]

heat of reaction from above value  comes  = -42.4 cal/gmole.

You should check your calculation.

 

Note however that a heat of reaction calculated from the standard heat of formation is only valid at 25 ^oC, and is based on solid P₂O₅, not gaseous..

It is not valid at 591 ^oC, or any other temperature that differs from 25 ^oC.

 

You should think hard what you really want to calculate.

[shamir]

Thanks for clarification.

 

 i want to share with your that our reaction is such that in reactor excess phosphorous is burn and converted to p2o5 vapour .During phosphorous combustion  in furnance it is converted to P2O5 Vapour . Also during combustion flame temperature reaches 845 Degree celcuis at point of combustion. Temperature of p2o5 reaches to 591 degree celcius and has to be reduced up to 60 0c . For this heat of reaction is needed.

[PingPong]

Temperature of p2o5 reaches to 591 degree celcius and has to be reduced up to 60 0c . For this heat of reaction is needed.

What reaction and what heat of reaction do you mean?

It seems to me you are asking how much heat is to be removed by cooling and condensing the vapor at 591 ^oC into solid at 60 ^oC, or am I missing something?

 

Do you have a simple flow scheme of the whole process?

What is the name and licensor of the process?

[shamir]

Yes Sir, I want to how to calculate heat to be removed for sub-cooling of vapor from 591 to 60 0c ,so i want your help.

Also i am sending u diagram for same 

Attached Files

•  model.pdf   14.28KB   20 downloads

[PingPong]

First note that P₂O₅ is generally assumed to be in reality P₄O₁₀ (also written as (P₂O₅)₂ or O₁₀P₄).

 

Note also that the mol weight of P₄O₁₀ is twice that of P₂O₅ so one has to be careful when using reported data per mol.

Now see the NIST-JANAF Thermochemical Table for crystal and gas P₄O₁₀ that I attached.

 

 P4O10 thermo data from JANAF.pdf   296.27KB   15 downloads

 

By interpolation of the data for P₄O_10(gas) you find that at 591 ^oC = 864 K the gas enthalpy (relative to gas at 298 K) is 148 kJ/mol and gas enthalpy at 60 ^oC = 333 K is 7 kJ/mol. See H^o - H^o(T_r) column in table.

 

Heat of sublimation of P₄O₁₀ at 60 ^oC is  104 kJ/mol (is heat of formation difference between gas and crystal at that temperature, see Δ_fH^o column in tables).

 

So total enthalpy difference between P₄O₁₀ gas at 591 ^oC and crystals at 60 ^oC is:

(148 - 7) + 104 = 245 kJ/mol P₄O₁₀ = 863 kJ/kg

 

In reality the transition between gas and crystal does not occur at 60 ^oC, but for the total enthalpy difference that does not matter. Nevertheless you can do the same exercise yourself by assuming that sublimation takes place at say 360 ^oC = 633 K. First use the gas table to obtain enthalpy difference between 591 and 360 ^oC; then determine heat of sublimation at 360 ^oC (from heats of formation); and then use crystal table to determine enthalpy difference between 360 to 60 ^oC. Add it all up and you should get practically the same value for the total enthalpy difference as I calculated before.

[shamir]

Thanks for solution. Also i have calculated heat load by knowing heat of reaction attached in below excel sheet. so please check and comment on the sheet .

 

Mean while i am doing calculation according to your suggestion .

Attached Files

•  Calculation sheet of Heat load.xlsx   10.94KB   36 downloads

[PingPong]

I am sorry to say that your calculation is full of mistakes.

 

Moreover I spent time to calculate that cooling P₂O₅ vapor from 591 to 60 oC gives a heat load of 863 kJ/kg = 206 kcal/kg but you completely ignore that as you seem to think that you can calculate the same with some avergae Cp. No you can't.

 

Apart from the calculation mistakes in the spreadsheet I have no idea what you are trying to calculate.

The spreadsheet ends with: Total heat load required for formation of P2O5 =F16+G34

I have no idea what "Total heat load required for formation of P2O5 " is supposed to mean, but adding up the contents of those two cells has no physical meaning whatsoever, even if the contensts of both cells were not wrong.

 

You have to clearly state what you want to calculate, and why.

You cannot just throw some numbers together, which were moreover calculated wrong.

[Art Montemayor]

Shamir:

 

There is nothing I can add to the expert advice and comments you have received from PingPong, a senior Forum member who has demonstrated his expertise in helping many of our members.  However, what I can do is attempt to expedite and stress the very important points PingPong has stated.  I have to assume that this is an existing, important industrial application and not an academic exercise and in keeping with that, I want to accentuate the importance of what PingPong has requested and emphasized:

• In order to receive the best and most accurate engineering recommendations from our members, you must furnish sufficient, logical, accurate, and detailed basic data.  Included in this information should be flow diagrams or P&IDs.  In the absence or inability to furnish these drawings a detailed sketch is at least some help.
• Clear, specific, and concise explanations are necessary to assist in an accurate and positive recommendation.  For example, you state “our reaction is such that in reactor excess phosphorous is burn and converted to p2o5 vapour”.  Is this correct?  Is the phosphorous in excess?  - or is the oxygen fed in excess to secure complete combustion?  This may or may not make a difference.  A detailed heat and mass balance in a flow diagram would be of significant help.
• You fail to address PingPong’s questions: “Do you have a simple flow scheme of the whole process?” and “What is the name and licensor of the process?”  It is not inconceivable that a senior engineer like PingPong may just know the process or have the experience with the licensor.  He, and all our other members, can only help you to the extent that you help them out with all your information.  You may not want - or you may not be able - to furnish detailed information; but at least let our members know that without leaving them hanging in the air waiting for details they will not receive.