# [SOLVED] enthalpy change

Printable View

• Dec 10th 2008, 02:49 PM
euclid2
[SOLVED] enthalpy change
Question from my grade 11 advanced chemistry class
The enthalpy change for the formation of two wolfram bromides are shown below.

$W_(s) + 2Br_2(l) \rightarrow WBr_4(s)
delta H_1 = -146.7 kJ$

$W_(s) + 3Br_2(l) \rightarrow WBr_6(s)
delta H_2 = -184.4 kJ$

Calculate the standard enthalpy change for the following reaction.

$Br_2(l) + WBr_4(s) \rightarrow WBr_6(s)$
• Dec 10th 2008, 03:40 PM
o_O
\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.
• Dec 10th 2008, 03:44 PM
euclid2
Quote:

Originally Posted by o_O
\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 $2Br_2$ ?
• Dec 10th 2008, 03:51 PM
o_O
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.
• Dec 10th 2008, 03:54 PM
euclid2
Quote:

Originally Posted by o_O
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?

$C_7H_14 + O_2 \rightarrow CO_2 + H_2O$
• Dec 10th 2008, 04:12 PM
o_O
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.
• Dec 10th 2008, 04:14 PM
euclid2
Quote:

Originally Posted by o_O
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