Anonymous
Anonymous asked in Education & ReferenceHomework Help · 9 years ago

Calculate the Delta H*rxn for the following @ 32degrees celcius?

Please help! SOS! My college chemistry professor has officially lost his mind and gave homework over the thanksgiving holiday!!!

A. 2CH4(g)=>C2H6(g)+H2(g)

B. 2NH3(g)=>N2H4(g)+H2(g)

C.2KClO3(s)=>2KCl(s)+3 O2(g)

Update:

Its do in two days!!!

Update 2:

it is online HW with no hints on how to go about answering the question

2 Answers

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  • Anonymous
    9 years ago
    Favorite Answer

    OK Tinkerbell... to solve these problems we will use two steps: (1) look up the delta H of formation of the compounds (which will be in a standard table at 25 degrees celcius) and then use Hess's equation to determine the delta H of the reaction desired, and (2) use heat capacities to take the result from (1) which is for 25 degrees C, and correct it for 32 degrees C.

    A.] 2CH4(g) => C2H6(g) + H2(g)

    First make sure it is a balanced reaction, and it is.

    Next look up the standard enthalpy of formation (at 25 degrees C) for each compound in the reaction

    using the websites below:

    Delta Hf for CH4(g) = - 74.9 kJ/mole

    Delta Hf for C2H6(g) = - 83.8 kJ/mole

    Delta Hf for H2(g) = 0.0 kJ/mole

    Note: You have to make sure the state of each compound (solid, liquid or gas) from the table is what you want, because it makes a difference.

    Now Hess's Law says for a given temperature:

    delta H for the reaction = (sum of the delta Hf of the products) - (sum of delta Hf of the reactants)

    So for this reaction: 2CH4(g) => C2H6(g) + H2(g) using the number of moles for each compound from the balanced reaction:

    delta H for the reaction = (1 mole of C2H6(g))(- 83.8 kJ/mole) + (1 mole of H2(g))(0.0 kJ/mole) - (2 moles of CH4(g))(- 74.9 kJ/mole) = 66.0 kJ/mole

    OK.. the delta H for this reaction = 66.0 kJ/mole is for the reaction at 25 degrees C

    Now we will use heat capacity data to compensate for raising the reaction temperature from 25 degrees C to 32 degrees C. From the websites below the heat capacities at constant pressure (1 atm.) of each compound in the gaseous state at 25 C below:

    Cp for CH4(g) = 0.0357 kJ/(mol K)

    Cp for C2H6(g) = 0.0525 kJ/(mol K)

    Cp for H2(g) = 0.0288 kJ/(mol K)

    delta Hf of reaction at 32 C = (delta Hf of reaction at 25 C) + (sum of Cp(delta T) of products) - (sum of Cp(delta T) of reactants), where delta T = 32C - 25 C = 7 C = 7 degrees K

    2CH4(g) => C2H6(g) + H2(g) using the number of moles for each compound from the balanced reaction:

    delta Hf of reaction at 32C = 66.0 kJ/mole + (1 mole of C2H6(g))(0.0525 kJ/(mole-K))(7 K) + (1 mole of H2(g))(0.0288 kJ/(moleK))(7 K) - (2 moles of CH4(g))(0.0357 kJ/(mole-K))(7 K) = 66.1 kJ/mole

    Note: as you can see the temperature change for this reaction has little affect on the reaction's delta H

    Now the other two problems should be done exactly the same way. I've attached the websites for the heats of formation of the compounds and their heat capacities.

    I'll leave the solutions for you to do.

    Here's a summary of the thermodynamic data for problems B and C:

    B.]

    Delta Hf for NH3(g) = - 45.9 kJ/mole

    Delta Hf for N2H4(g) = 95.4 kJ/mole

    Delta Hf for H2(g) = 0.0 kJ/mole

    Cp for NH3(g) = 0.0351 kJ/(mol K)

    Cp for N2H4(g) = 0.0602 kJ/(mol K)

    Cp for H2(g) = 0.0288 kJ/(mol K)

    C.]

    Delta Hf for KClO3(s) = - 391.4 kJ/mole

    Delta Hf for KCl(s) = - 436.7 kJ/mole

    Delta Hf for O2(g) = 0.0 kJ/mole

    Cp for KClO3(s) = 0.1075 kJ/(mol K)

    Cp for KCl(s) = 0.0506 kJ/(mol K)

    Cp for O2(g) = 0.0294 kJ/(mol K)

    Note: KClO3 and KCl are in the solid state

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  • Sherri
    Lv 4
    4 years ago

    Okay, this is related to Hess' Law which states that the enthalpy change of a reaction is independent of the path taken. The standard enthalpy of formation of a compound is the heat energy released when one mole of the compound is formed from its constituent elements in their standard physical states under standard conditions. You could imagine the reactants on the left hand side of an equation being deconstructed into their constituent elements in their standard physical states(heat energy is absorbed when this happens) and the product(s) on the right hand side of that equation being formed from their constituent elements(heat energy is released when this happens). Having visualised this, it would make sense that we subtract the sum of the standard enthalpies of formation of the reactants from those of the products. (1) ΔHf(C₅H₁₂)-5ΔHf(CO₂)-6ΔHf(H₂O)= -3505.8kJ (2) ΔHf(CO₂)= 393.5kJ --->ΔHf(C) and ΔHf(O₂) are 0kJ since carbon and oxygen are in their standard physical states (3) ΔHf(H₂O)= 285.83 -->I believe that there's an error in the information provided since it is the ΔHf of liquid water that is needed, not that of gaseous water. I checked the ΔHf values of all the reactants and it appears that the values only tally if the ΔHf of liquid water is used. ΔHf(C₅H₁₂)-5(393.5kJ/mol)-6(285.83kJ/m... -3505.8kJ ΔHf(C₅H₁₂)= -3505.8kJ+5(393.5kJ/mol)+6(285.83kJ/mol) = 176.68kJ/mol Since C and H₂ are in their standard physical states, ΔHrxn = -ΔHf(C₅H₁₂) = -176.68kJ I hope this helps and feel free to send me an e-mail if you have any doubts!

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