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Math Help - [SOLVED] enthalpy change

  1. #1
    Senior Member euclid2's Avatar
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    [SOLVED] enthalpy change

    Question from my grade 11 advanced chemistry class
    The enthalpy change for the formation of two wolfram bromides are shown below.

     W_(s) + 2Br_2(l) \rightarrow WBr_4(s) <br />
             delta H_1 = -146.7 kJ
     W_(s) + 3Br_2(l) \rightarrow WBr_6(s)<br />
             delta H_2 = -184.4 kJ

    Calculate the standard enthalpy change for the following reaction.

     Br_2(l) + WBr_4(s) \rightarrow WBr_6(s)
    Last edited by euclid2; December 10th 2008 at 02:07 PM.
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  2. #2
    o_O
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    \begin{aligned} \text{W}_{(s)}  + 2\text{Br}_{2 (l)} & \rightarrow \text{WBr}_{4 (s)} & \Delta H_1 = -146.7 \ \text{kJ} \\ \text{W}_{(s)}  + 3\text{Br}_{2 (l)} & \rightarrow \text{WBr}_{6(s)} & \Delta H_2 = -184.4 \ \text{kJ} \end{aligned}

    Reverse the first reaction:

    \begin{aligned} \text{WBr}_{4 (s)} & \rightarrow \text{W}_{(s)}  + 2\text{Br}_{2 (l)} & \Delta H_1 = {\color{red}+}146.7 \ \text{kJ} \\ \text{W}_{(s)}  + 3\text{Br}_{2 (l)} & \rightarrow \text{WBr}_{6(s)} & \Delta H_2 = -184.4 \ \text{kJ} \end{aligned}

    Now add the two reactions together. Note which intermediates get excluded and you will get your net reaction and the required enthalpy change.
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    Senior Member euclid2's Avatar
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    Quote Originally Posted by o_O View Post
    \begin{aligned} \text{W}_{(s)}  + 2\text{Br}_{2 (l)} & \rightarrow \text{WBr}_{4 (s)} & \Delta H_1 = -146.7 \ \text{kJ} \\ \text{W}_{(s)}  + 3\text{Br}_{2 (l)} & \rightarrow \text{WBr}_{6(s)} & \Delta H_2 = -184.4 \ \text{kJ} \end{aligned}

    Reverse the first reaction:

    \begin{aligned} \text{WBr}_{4 (s)} & \rightarrow \text{W}_{(s)}  + 2\text{Br}_{2 (l)} & \Delta H_1 = {\color{red}+}146.7 \ \text{kJ} \\ \text{W}_{(s)}  + 3\text{Br}_{2 (l)} & \rightarrow \text{WBr}_{6(s)} & \Delta H_2 = -184.4 \ \text{kJ} \end{aligned}

    Now add the two reactions together. Note which intermediates get excluded and you will get your net reaction and the required enthalpy change.
    I got this far, Although how do I end up with  Br_2 from the remaining 3Br_2 and  2Br_2 ?
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    o_O
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    3 \text{Br}_2 is used up and 2 \text{Br}_2 is produced. Looking a the net reaction, this is the same thing as saying 1 Br_2 is used up.
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    Senior Member euclid2's Avatar
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    Quote Originally Posted by o_O View Post
    3 \text{Br}_2 is used up and 2 \text{Br}_2 is produced. Looking a the net reaction, this is the same thing as saying 1 Br_2 is used up.
    Thanks for the help. While were talking chemistry, do you know how to write a balanced equation for the combustion of cycloheptane other then putting  10 \frac{1}{2} infront of  O_2 in the reactant?

     C_7H_14 + O_2 \rightarrow CO_2 + H_2O
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    o_O
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    It's just a matter of logic really. Also, sticking that 10 \tfrac{1}{2} wouldn't balance the equation at all since we haven't compensated for the carbons and hydrogens!

    Notice that there's 7 carbons and 14 hydrogens on the left. We can stick a 7 in front of carbon dioxide to balance the carbons (it's our only source!) and stick a 7 in front of our dihydrogen monoxide (since this is our only source as well).

    Now, let's compensate for the oxygens. On the right hand side, we would have 14 oxygens from \text{CO}_2 and 7 oxygens from \text{H}_2\text{O} giving us a total of 21 oxygens.

    This suggests sticking in \tfrac{21}{2} in front of O_2 on the left hand side and we'll be done.

    \text{C}_7\text{H}_{14} + {\color{red}\tfrac{21}{2}}\text{O}_2 \rightarrow {\color{red}7}\text{CO}_2 + {\color{red}7}\text{H}_2\text{O}

    If you don't like fractions, multiply both sides by 2 and you'll get integer values.
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  7. #7
    Senior Member euclid2's Avatar
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    Quote Originally Posted by o_O View Post
    It's just a matter of logic really. Also, sticking that 10 \tfrac{1}{2} wouldn't balance the equation at all since we haven't compensated for the carbons and hydrogens!

    Notice that there's 7 carbons and 14 hydrogens on the left. We can stick a 7 in front of carbon dioxide to balance the carbons (it's our only source!) and stick a 7 in front of our dihydrogen monoxide (since this is our only source as well).

    Now, let's compensate for the oxygens. On the right hand side, we would have 14 oxygens from \text{CO}_2 and 7 oxygens from \text{H}_2\text{O} giving us a total of 21 oxygens.

    This suggests sticking in \tfrac{21}{2} in front of O_2 on the left hand side and we'll be done.

    \text{C}_7\text{H}_{14} + {\color{red}\tfrac{21}{2}}\text{O}_2 \rightarrow {\color{red}7}\text{CO}_2 + {\color{red}7}\text{H}_2\text{O}

    If you don't like fractions, multiply both sides by 2 and you'll get integer values.
    Absolutely, I did have those 7`s there, I just forgot to mention that. Super Member! sweet
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