Enthalpy Change

  • Thread starter erik05
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Hello. Could anyone help me out with a question I'm stuck on? Thanks in advance.


Calcium metal reacts with hydrochloric acid in a calorimeter and the enthalpy change is determined. Similarly, the enthalpy change for the reaction of calcium oxide with hydrochloric acid is determined. What is the enthalpy change for the following reaction:

[tex] Ca_{(s)}+ 1/2 O_{2} \longrightarrow CaO_{(s)} [/tex]


Evidence:
[tex] Ca_{(s)} + 2HCl_{(aq)} \longrightarrow CaCl_{2}_{(aq)} + H_{2}_{(g)} ,\triangle H_{r}=? [/tex]

[tex] CaO_{(s)} + 2HCl_{(aq)} \longrightarrow CaCl_{2}_{(aq)} + H_{2}O_{(l)} ,\triangle H_{r}=? [/tex]

Temperature Change Due to Reaction With 100 mL of 1.0 mol/L HCl

Reactant: Ca(s)
Mass(g) : 0.52
Initial Temp: 21.3 °C
Final Temp: 34.5 °C


Reactant: CaO(s)
Mass (g): 1.47
Initial Temp: 21.1 °C
Final Temp: 28.0 °C

So far I have:

for Ca:
[tex] nH= mc\triangle t [/tex]
[tex] H= \frac{mc\triangle t}{n} [/tex]
H= {(0.100 L)*(1.0 mol/1 L)*(0.03646 kg/1 mol)*(4.19 kJ/kg°C)*(34.5°C-21.3°C)} / { (0.52g)*(1 mol/40.08 g) } = - 15.54 kJ/mol

for CaO:
H= {(0.100 L)*(1.0 mol/1 L)*(0.03646 kg/1 mol)*(4.19 kJ/kg°C)*(28.0°C-21.1°C)} / { (1.47g)*(1 mol/56.08 g) } = -4.02 kJ/mol

Not too sure where to go from here but the answers are suppose to be -0.43 MJ,+ 0.11 MJ and -0.60 MJ,also I'm not too sure if that's the order the answers appear either.
 

Answers and Replies

  • #2
xanthym
Science Advisor
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erik05 said:
Hello. Could anyone help me out with a question I'm stuck on? Thanks in advance.


Calcium metal reacts with hydrochloric acid in a calorimeter and the enthalpy change is determined. Similarly, the enthalpy change for the reaction of calcium oxide with hydrochloric acid is determined. What is the enthalpy change for the following reaction:

[tex] Ca_{(s)}+ 1/2 O_{2} \longrightarrow CaO_{(s)} [/tex]


Evidence:
[tex] Ca_{(s)} + 2HCl_{(aq)} \longrightarrow CaCl_{2}_{(aq)} + H_{2}_{(g)} ,\triangle H_{r}=? [/tex]

[tex] CaO_{(s)} + 2HCl_{(aq)} \longrightarrow CaCl_{2}_{(aq)} + H_{2}O_{(l)} ,\triangle H_{r}=? [/tex]

Temperature Change Due to Reaction With 100 mL of 1.0 mol/L HCl

Reactant: Ca(s)
Mass(g) : 0.52
Initial Temp: 21.3 °C
Final Temp: 34.5 °C


Reactant: CaO(s)
Mass (g): 1.47
Initial Temp: 21.1 °C
Final Temp: 28.0 °C

So far I have:

for Ca:
[tex] nH= mc\triangle t [/tex]
[tex] H= \frac{mc\triangle t}{n} [/tex]
H= {(0.100 L)*(1.0 mol/1 L)*(0.03646 kg/1 mol)*(4.19 kJ/kg°C)*(34.5°C-21.3°C)} / { (0.52g)*(1 mol/40.08 g) } = - 15.54 kJ/mol

for CaO:
H= {(0.100 L)*(1.0 mol/1 L)*(0.03646 kg/1 mol)*(4.19 kJ/kg°C)*(28.0°C-21.1°C)} / { (1.47g)*(1 mol/56.08 g) } = -4.02 kJ/mol

Not too sure where to go from here but the answers are suppose to be -0.43 MJ,+ 0.11 MJ and -0.60 MJ,also I'm not too sure if that's the order the answers appear either.
Your mistake in computing ΔH for the Ca & CaO reactions was in using HCl mass in the ΔH equations. The WATER mass in each reaction is required for the ΔH equations since it's primarily WATER that is being heated and is changing temperature.
For Ca:
ΔH = -{(100 mL)*(10(-3) kg/ml)*(4.19 kJ/kg°C)*(34.5°C - 21.3°C)/{(0.52 g)*(1 mol/40.08 g)} =
= (-426.3 kJ/mol)

For CaO:
ΔH = -{(100 mL)*(10(-3) kg/ml)*(4.19 kJ/kg°C)*(28.0°C - 21.1°C)}/{(1.47 g)*(1 mol/56.08 g) } =
= (-110.3 kJ/mol)

To determine ΔH for the subject reaction, we form its reaction equation by adding reaction equations with known ΔH (same entities on both sides of arrow cancel):

[tex] Ca_{(s)} \ + \ 2HCl_{(aq)} \ \longrightarrow \ CaCl_{2}_{(aq)} \ + \ H_{2}_{(g)} \ \ \ \ \color{red} \triangle H_{r}=(-426.3 \ kJ)=(-0.43 \ MJ) [/tex]

[tex] CaCl_{2}_{(aq)} \ + \ H_{2}O_{(l)} \ \longrightarrow \ CaO_{(s)} \ + \ 2HCl_{(aq)} \ \ \ \ \color{red} \triangle H_{r}=(+110.3 \ kJ)=(+0.11 \ MJ) [/tex]

[tex] H_{2}_{(g)} \ + \ (1/2) O_{2} \ \longrightarrow \ H_{2}O_{(l)} \ \ \ \ \color{red} \triangle H_{r}=(-286 \ kJ)=(-0.28 \ MJ) [/tex]

----------------------------------------------------------

[tex] Ca_{(s)} \ + \ (1/2) O_{2} \ \longrightarrow \ CaO_{(s)} \ \ \ \ \color{red} \triangle H_{r}=(-602 \ kJ)=(-0.60 \ MJ) [/tex]


~~
 
Last edited:
  • #3
50
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Ah, I understand now. Thank you!
 

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