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Thread: [SOLVED] enthalpy change

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    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.

    $\displaystyle W_(s) + 2Br_2(l) \rightarrow WBr_4(s)
    delta H_1 = -146.7 kJ$
    $\displaystyle W_(s) + 3Br_2(l) \rightarrow WBr_6(s)
    delta H_2 = -184.4 kJ$

    Calculate the standard enthalpy change for the following reaction.

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

    Reverse the first reaction:

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

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

    Reverse the first reaction:

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

    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 $\displaystyle Br_2 $ from the remaining $\displaystyle 3Br_2 $ and $\displaystyle 2Br_2 $ ?
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    3 $\displaystyle \text{Br}_2$ is used up and 2 $\displaystyle \text{Br}_2$ is produced. Looking a the net reaction, this is the same thing as saying 1 $\displaystyle Br_2$ 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 $\displaystyle \text{Br}_2$ is used up and 2 $\displaystyle \text{Br}_2$ is produced. Looking a the net reaction, this is the same thing as saying 1 $\displaystyle Br_2$ 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 $\displaystyle 10 \frac{1}{2} $ infront of $\displaystyle O_2 $ in the reactant?

    $\displaystyle C_7H_14 + O_2 \rightarrow CO_2 + H_2O $
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    It's just a matter of logic really. Also, sticking that $\displaystyle 10 \tfrac{1}{2}$ 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 $\displaystyle \text{CO}_2$ and 7 oxygens from $\displaystyle \text{H}_2\text{O}$ giving us a total of 21 oxygens.

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

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

    If you don't like fractions, multiply both sides by 2 and you'll get integer values.
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    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 $\displaystyle 10 \tfrac{1}{2}$ 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 $\displaystyle \text{CO}_2$ and 7 oxygens from $\displaystyle \text{H}_2\text{O}$ giving us a total of 21 oxygens.

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

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

    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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