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Use the given information to determine the enthalpy of the reaction

Jean Blumer
2022-01-10
Answered

1. $2M\left(s\right)+6HCl\left(aq\right)\Rightarrow 2MCl3\left(aq\right)+3H2\left(g\right)2M\left(s\right)+6HCl\left(aq\right)\Rightarrow 2MCl3\left(aq\right)+3H2\left(g\right)$

$\mathrm{\u25b3}{H}_{1}=-819.0kJ$

2.$HCl\left(g\right)\Rightarrow HCl\left(aq\right)HCl\left(g\right)\Rightarrow HCl\left(aq\right)$

$\mathrm{\u25b3}{H}_{2}=-74.8kJ$

3.$H2\left(g\right)+Cl2\left(g\right)\Rightarrow 2HCl\left(g\right)H2\left(g\right)+Cl2\left(g\right)\Rightarrow 2HCl\left(g\right)$

$\mathrm{\u25b3}{H}_{3}=-1845.0kJ$

4.$MCl3\left(s\right)\Rightarrow MCl3\left(aq\right)$

$\mathrm{\u25b3}{H}_{4}=-128.0kJ$

Use the given information to determine the enthalpy of the reaction

$2M\left(s\right)+3C{l}_{2}\left(g\right)\Rightarrow 2MC{l}_{3}\left(s\right)$

$\mathrm{\u25b3}H=?$

2.

3.

4.

Use the given information to determine the enthalpy of the reaction

You can still ask an expert for help

Melinda McCombs

Answered 2022-01-11
Author has **38** answers

Step 1

In the above reactions we only need 2 reactions to calculate the enthalpy of the following reaction;

$2M\left(s\right)+3Cl2\left(g\right)\Rightarrow MCl3\left(g\right)$

Step 2

We have 4 chemical equations from which 2 are important:

$2M\left(s\right)+6HCl\left(aq\right)\Rightarrow 2MC{l}_{3}\left(aq\right)+3{H}_{2}\left(g\right)$

$\mathrm{\u25b3}{H}_{1}=819KJ$

${H}_{2}\left(g\right)+C{l}_{2}\left(g\right)\Rightarrow 2HCl\left(g\right)$

$\mathrm{\u25b3}{H}_{3}=-1845KJ$

Multiplying equation 1st by 3

$3{H}_{2}\left(g\right)+3C{l}_{2}\left(g\right)\Rightarrow 6HCl\left(g\right)$

$\mathrm{\u25b3}{H}_{3}=3\times (-1845KJ)=-5535KJ$

Adding equation 1st and 3rd

$2M\left(s\right)+3C{l}_{2}\left(g\right)\Rightarrow 2MC{l}_{3}\left(aq\right)$

$\mathrm{\u25b3}{H}_{reaction}$

$=(\mathrm{\u25b3}{H}_{1}+\mathrm{\u25b3}{H}_{3})=-4716KJ$

Therefore the enthalpy of formation$=-4716KJ$ (answer)

In the above reactions we only need 2 reactions to calculate the enthalpy of the following reaction;

Step 2

We have 4 chemical equations from which 2 are important:

Multiplying equation 1st by 3

Adding equation 1st and 3rd

Therefore the enthalpy of formation

Ben Owens

Answered 2022-01-12
Author has **27** answers

Chemical reaction 1: $2M\left(s\right)+6HCl\left(aq\right)\Rightarrow 2MCl3\left(aq\right)+3H2\left(g\right);\mathrm{\u25b3}H1=-556.0kJ$ . Chemical reaction 2: $HCl\left(g\right)\Rightarrow HCl\left(aq\right);\mathrm{\u25b3}H2=-74.8kJ$ .

Chemical reaction 3:$H2\left(g\right)+Cl2\left(g\right)\Rightarrow 2HCl\left(g\right);\mathrm{\u25b3}H3=-1845.0kJ$

Chemical reaction 4:$MCl3\left(s\right)\Rightarrow MCl3\left(aq\right);\mathrm{\u25b3}H4=-342.0kJ$ .

Chemical reaction 5:$2M\left(s\right)+3Cl2\left(g\right)\Rightarrow 2MCl3\left(s\right);\mathrm{\u25b3}H5=?.\mathrm{\u25b3}H5=\mathrm{\u25b3}H1+6\xb7\mathrm{\u25b3}H2+3\cdot \mathrm{\u25b3}H3-2\xb7\mathrm{\u25b3}H4$ .

$\mathrm{\u25b3}H5=-550kJ+6\xb7(-74,8kJ)+3\cdot (-1845kJ)-2\cdot (-342kJ)$ .

$\mathrm{\u25b3}H5=-550kJ-448,8kJ-5535kJ+684kJ$ .

$\mathrm{\u25b3}H5=-5849,8kJ$ .

Chemical reaction 3:

Chemical reaction 4:

Chemical reaction 5:

star233

Answered 2022-01-15
Author has **137** answers

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