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    given the data:
    4nh3(g) + 302(g) -->2n2(g) + 6h20(l), dletah =-1530kjmol-1
    h2(g) + 0.5o2 (g) -->h2o(l), delta h =-288kjmol-1
    calculate the entalpy of formation of ammonia
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    not sure exactly but its enthalpy change of formation so arrows go upwards. 1 mole of each product should form so i think because formation of h20 is =-288 you have to divide that value by 6 something like that.
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    (Original post by nina313)
    given the data:
    4nh3(g) + 302(g) -->2n2(g) + 6h20(l), dletah =-1530kjmol-1
    h2(g) + 0.5o2 (g) -->h2o(l), delta h =-288kjmol-1
    calculate the entalpy of formation of ammonia.
    The correct answer is -152Kjmol-1 but i dont know how to solve it?
    Your first step is to write out the equation for the enthalpy of formation of ammonia.

    Then see if you can construct this equation using the equations given. Make sure yo do the same to the energy change as to each operation you carry out on each equation.
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    Name:  Screenshot (198).png
Views: 8
Size:  11.8 KB
    What am I doing wrong?
    please help!


    (Original post by charco)
    Your first step is to write out the equation for the enthalpy of formation of ammonia.

    Then see if you can construct this equation using the equations given. Make sure yo do the same to the energy change as to each operation you carry out on each equation.
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    (Original post by mansnothot)
    not sure exactly but its enthalpy change of formation so arrows go upwards. 1 mole of each product should form so i think because formation of h20 is =-288 you have to divide that value by 6 something like that.
    I thinks its multiply by 6 and not divide by 6
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    (Original post by nina313)
    Name:  Screenshot (198).png
Views: 8
Size:  11.8 KB
    What am I doing wrong?
    please help!
    Your equation:

    4NH3 + 3O2 --> 2N2 + 6H2O ..... ΔH = -1530 kJ

    This is equal to -(4 * ΔHf(NH3) + (6 * ΔHf(H2O)

    -1530 - ( 6 x -288) = -(4 * ΔHf(NH3)

    -1530 + 1728 = -(4 * ΔHf(NH3)

    ΔHf(NH3) = -198/4 = -49.5 kJ
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    (Original post by charco)
    Your equation:

    4NH3 + 3O2 --> 2N2 + 6H2O ..... ΔH = -1530 kJ

    This is equal to -(4 * ΔHf(NH3) + (6 * ΔHf(H2O)

    -1530 - ( 6 x -288) = -(4 * ΔHf(NH3)

    -1530 + 1728 = -(4 * ΔHf(NH3)

    ΔHf(NH3) = -198/4 = -49.5 kJ
    Thankyou
 
 
 
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