# 1. 2M(s)+6HCl(aq) \Rightarrow 2MCl3(aq)+3H2(g)2M(s)+6HCl(aq) \Rightarrow 2MCl3(aq)+3H2(g) \triangle H_{1}=-819.0kJ 2. HCl(g) \Rightarrow HCl(aq)HCl(g)

1. $2M\left(s\right)+6HCl\left(aq\right)⇒2MCl3\left(aq\right)+3H2\left(g\right)2M\left(s\right)+6HCl\left(aq\right)⇒2MCl3\left(aq\right)+3H2\left(g\right)$
$\mathrm{△}{H}_{1}=-819.0kJ$
2. $HCl\left(g\right)⇒HCl\left(aq\right)HCl\left(g\right)⇒HCl\left(aq\right)$
$\mathrm{△}{H}_{2}=-74.8kJ$
3. $H2\left(g\right)+Cl2\left(g\right)⇒2HCl\left(g\right)H2\left(g\right)+Cl2\left(g\right)⇒2HCl\left(g\right)$
$\mathrm{△}{H}_{3}=-1845.0kJ$
4. $MCl3\left(s\right)⇒MCl3\left(aq\right)$
$\mathrm{△}{H}_{4}=-128.0kJ$
Use the given information to determine the enthalpy of the reaction
$2M\left(s\right)+3C{l}_{2}\left(g\right)⇒2MC{l}_{3}\left(s\right)$
$\mathrm{△}H=?$
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Melinda McCombs
Step 1
In the above reactions we only need 2 reactions to calculate the enthalpy of the following reaction;
$2M\left(s\right)+3Cl2\left(g\right)⇒MCl3\left(g\right)$
Step 2
We have 4 chemical equations from which 2 are important:
$2M\left(s\right)+6HCl\left(aq\right)⇒2MC{l}_{3}\left(aq\right)+3{H}_{2}\left(g\right)$
$\mathrm{△}{H}_{1}=819KJ$
${H}_{2}\left(g\right)+C{l}_{2}\left(g\right)⇒2HCl\left(g\right)$
$\mathrm{△}{H}_{3}=-1845KJ$
Multiplying equation 1st by 3
$3{H}_{2}\left(g\right)+3C{l}_{2}\left(g\right)⇒6HCl\left(g\right)$
$\mathrm{△}{H}_{3}=3×\left(-1845KJ\right)=-5535KJ$
$2M\left(s\right)+3C{l}_{2}\left(g\right)⇒2MC{l}_{3}\left(aq\right)$
$\mathrm{△}{H}_{reaction}$
$=\left(\mathrm{△}{H}_{1}+\mathrm{△}{H}_{3}\right)=-4716KJ$
Therefore the enthalpy of formation $=-4716KJ$ (answer)
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Ben Owens
Chemical reaction 1: $2M\left(s\right)+6HCl\left(aq\right)⇒2MCl3\left(aq\right)+3H2\left(g\right);\mathrm{△}H1=-556.0kJ$. Chemical reaction 2: $HCl\left(g\right)⇒HCl\left(aq\right);\mathrm{△}H2=-74.8kJ$.
Chemical reaction 3: $H2\left(g\right)+Cl2\left(g\right)⇒2HCl\left(g\right);\mathrm{△}H3=-1845.0kJ$
Chemical reaction 4: $MCl3\left(s\right)⇒MCl3\left(aq\right);\mathrm{△}H4=-342.0kJ$.
Chemical reaction 5: $2M\left(s\right)+3Cl2\left(g\right)⇒2MCl3\left(s\right);\mathrm{△}H5=?.\mathrm{△}H5=\mathrm{△}H1+6·\mathrm{△}H2+3\cdot \mathrm{△}H3-2·\mathrm{△}H4$.
$\mathrm{△}H5=-550kJ+6·\left(-74,8kJ\right)+3\cdot \left(-1845kJ\right)-2\cdot \left(-342kJ\right)$.
$\mathrm{△}H5=-550kJ-448,8kJ-5535kJ+684kJ$.
$\mathrm{△}H5=-5849,8kJ$.
###### Not exactly what you’re looking for?
star233

$\phantom{\rule{0ex}{0ex}}2M\left(s\right)+3C{l}_{2}\left(g\right)⇒2MC{l}_{3}\left(s\right)\phantom{\rule{0ex}{0ex}}\mathrm{△}H=-944kJ+6×\left(-74.8kJ\right)+3×\left(-1845kJ\right)+\left(2\right)×\left(234kJ\right)\phantom{\rule{0ex}{0ex}}\mathrm{△}H=-944kJ-448.8-5535kJ+468kJ\phantom{\rule{0ex}{0ex}}\mathrm{△}H=-6459.8kJ$