Consider these reactions, where M represents a generic metal.
2M(s)+6HCl(aq)⟶2MCl3(aq)+3H2(g)Δ𝐻1=−901.0 kJ
HCl(g)⟶HCl(aq) Δ𝐻2=−74.8 kJ
H2(g)+Cl2(g)⟶2HCl(g) Δ𝐻3=−1845.0 kJ
MCl3(s)⟶MCl3(aq) Δ𝐻4=−195.0 kJ
Use the given information to determine the enthalpy of the reaction
2M(s)+3Cl2(g)⟶2MCl3(s)
To determine the enthalpy of the reaction 2M(s) + 3Cl2(g) ⟶ 2MCl3(s), we can use Hess's Law.
Hess's Law states that the enthalpy change of a reaction is independent of the pathway taken and depends only on the initial and final states of the reaction. Therefore, we can manipulate the given reactions to obtain the desired reaction and then sum up their enthalpy changes.
First, we need to cancel out any common compounds that appear on both sides of the equations. We can cancel out HCl(aq) from reaction 1 and HCl(aq) from reaction 4, leaving us with:
2M(s) + 6HCl(aq) ⟶ 2MCl3(aq) + 3H2(g) (equation 1)
HCl(g) ⟶ HCl(aq) (equation 2)
H2(g) + Cl2(g) ⟶ 2HCl(g) (equation 3)
MCl3(s) ⟶ MCl3(aq) (equation 4)
Next, we can manipulate these equations to obtain the desired reaction:
Multiply equation 1 by -1:
-2M(s) - 6HCl(aq) ⟶ -2MCl3(aq) - 3H2(g)
Multiply equation 2 by -3:
-3HCl(g) ⟶ -3HCl(aq)
Multiply equation 3 by 2:
2H2(g) + 2Cl2(g) ⟶ 4HCl(g)
Finally, we can sum up these equations to obtain our desired reaction:
-2M(s) - 3HCl(aq) - 3HCl(g) + 2H2(g) + 2Cl2(g) ⟶ -2MCl3(s) - 2MCl3(aq)
Now, we can sum up the enthalpy changes of these equations to determine the enthalpy of the desired reaction:
ΔH = ΔH1 + ΔH2 + ΔH3 + ΔH4
ΔH = -901.0 kJ + (-74.8 kJ) + (-1845.0 kJ) + (-195.0 kJ)
ΔH = -901.0 kJ - 74.8 kJ - 1845.0 kJ - 195.0 kJ
ΔH = -3015.8 kJ
Therefore, the enthalpy of the reaction 2M(s) + 3Cl2(g) ⟶ 2MCl3(s) is -3015.8 kJ.
To determine the enthalpy of the reaction
2M(s) + 3Cl2(g) ⟶ 2MCl3(s),
we can use Hess's law, which states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual steps in the reaction.
Let's break down the reaction into smaller steps using the given information:
Step 1: 2M(s) + 6HCl(aq) ⟶ 2MCl3(aq) + 3H2(g)
Hess's Law allows us to manipulate the given reactions to match the desired reaction. We need to reverse ΔH1 as well as multiply it by a factor of 2 to match the coefficients in the desired reaction. So the enthalpy change for Step 1 is:
2 × (ΔH1) = 2 × (-901.0 kJ) = -1802.0 kJ
Step 2: HCl(g) ⟶ HCl(aq)
This reaction is given as ΔH2. We don't need to manipulate it as it already matches the desired reaction.
ΔH2 = -74.8 kJ
Step 3: H2(g) + Cl2(g) ⟶ 2HCl(g)
For Step 3, we need to reverse ΔH3 and change the sign to match the desired reaction.
-ΔH3 = -(-1845.0 kJ) = 1845.0 kJ
Step 4: MCl3(s) ⟶ MCl3(aq)
For Step 4, we also need to reverse ΔH4 and change the sign.
-ΔH4 = -(-195.0 kJ) = 195.0 kJ
Now, we can add up the enthalpy changes for the individual steps to get the overall enthalpy change for the desired reaction:
Overall enthalpy change = Step 1 + Step 2 + Step 3 + Step 4
= -1802.0 kJ + (-74.8 kJ) + 1845.0 kJ + 195.0 kJ
= 163.2 kJ
Therefore, the enthalpy of the reaction
2M(s) + 3Cl2(g) ⟶ 2MCl3(s)
is 163.2 kJ.
In order to determine the enthalpy change for the reaction 2M(s) + 3Cl2(g) ⟶ 2MCl3(s), we can use the given information and apply the Hess's Law.
Step 1: Start with the given reactions and their enthalpy changes:
1. 2M(s) + 6HCl(aq) ⟶ 2MCl3(aq) + 3H2(g) Δ𝐻1 = -901.0 kJ
2. HCl(g) ⟶ HCl(aq) Δ𝐻2 = -74.8 kJ
3. H2(g) + Cl2(g) ⟶ 2HCl(g) Δ𝐻3 = -1845.0 kJ
4. MCl3(s) ⟶ MCl3(aq) Δ𝐻4 = -195.0 kJ
Step 2: Rearrange the given reactions to match the desired reaction:
- Multiply reaction 1 by 2 to match the coefficients of MCl3:
4M(s) + 12HCl(aq) ⟶ 4MCl3(aq) + 6H2(g) Δ𝐻' = 2 × Δ𝐻1 = -1802.0 kJ
- Multiply reaction 2 by 2 to match the coefficients of HCl:
2HCl(g) ⟶ 2HCl(aq) Δ𝐻'' = 2 × Δ𝐻2 = -149.6 kJ
- Multiply reaction 3 by 2 to match the coefficients of HCl:
2H2(g) + 2Cl2(g) ⟶ 4HCl(g) Δ𝐻''' = 2 × Δ𝐻3 = -3690.0 kJ
Step 3: Combine the modified reactions to obtain the desired reaction:
Add the modified reactions together to cancel out common substances:
4M(s) + 12HCl(aq) + 2HCl(g) + 2H2(g) + 2Cl2(g) ⟶ 4MCl3(aq) + 6H2(g) + 4HCl(g)
Canceling out the common substances gives us:
4M(s) + 14HCl(aq) + 2Cl2(g) ⟶ 4MCl3(aq) + 4HCl(g)
Step 4: Sum up the enthalpy changes:
Add up the enthalpy changes of the modified reactions:
Δ𝐻_total = Δ𝐻' + Δ𝐻'' + Δ𝐻'''
Δ𝐻_total = -1802.0 kJ + (-149.6 kJ) + (-3690.0 kJ) = -4641.6 kJ
Therefore, the enthalpy change for the reaction 2M(s) + 3Cl2(g) ⟶ 2MCl3(s) is -4641.6 kJ.