H2O(g) + Cl2(g) 2 HCl(g) + ½ O2(g)
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COCl2(g) + H2O(g) CH2Cl2(l) + O2(g) DeltaH = 16 kJ
HCl(g) ½ H2(g) + ½ Cl2(g) Delta H = 92 kJ
CH2Cl2(l) + H2(g) + 3/2 O2(g) COCl2(g) + 2 H2O(g) DeltaH = -167 kJ
use hess's law
To solve this problem using Hess's Law, we need to manipulate the given equations and their corresponding enthalpy values to obtain the desired reaction equation. Here's how we can do that:
1. Reverse the direction of the second equation (HCl(g) = ½ H2(g) + ½ Cl2(g)), as we need HCl on the reactant side, not the product side. This will give us the equation: ½ H2(g) + ½ Cl2(g) = HCl(g) with a ΔH value of -92 kJ.
2. Multiply the second equation by 2 to obtain the equation: H2(g) + Cl2(g) = 2 HCl(g) with a ΔH value of -184 kJ. This ensures that the stoichiometric coefficients match the desired reaction equation.
3. Multiply the third equation by 2 to obtain: 2 CH2Cl2(l) + 2 H2(g) + 3 O2(g) = 2 COCl2(g) + 4 H2O(g) with a ΔH value of -334 kJ. This ensures that the stoichiometric coefficients match the desired reaction equation.
4. Add the first equation (COCl2(g) + H2O(g) = CH2Cl2(l) + O2(g)) to the second equation (H2(g) + Cl2(g) = 2 HCl(g)) to obtain: COCl2(g) + H2O(g) + H2(g) + Cl2(g) = CH2Cl2(l) + O2(g) + 2 HCl(g) with a ΔH value of -168 kJ.
5. Add the fourth equation (2 CH2Cl2(l) + 2 H2(g) + 3 O2(g) = 2 COCl2(g) + 4 H2O(g)) to the equation obtained in the previous step to obtain the desired reaction equation: H2O(g) + Cl2(g) = 2 HCl(g) + ½ O2(g) with a ΔH value of -502 kJ (sum of -168 kJ and -334 kJ).
Therefore, the heat of reaction for the given equation H2O(g) + Cl2(g) = 2 HCl(g) + ½ O2(g) is -502 kJ.
To find the heat of reaction for the given equation using Hess's Law, we need to combine the given equations to cancel out the common compounds and obtain the desired equation.
First, we will reverse the third equation:
COCl2(g) + 2 H2O(g) CH2Cl2(l) + H2(g) + 3/2 O2(g) DeltaH = 167 kJ
Then, we will multiply the second equation by 2 to match the number of moles of HCl:
2 HCl(g) H2(g) + Cl2(g) Delta H = 2 * 92 kJ = 184 kJ
Next, we will multiply the first equation by 2 to match the number of moles of O2:
2 COCl2(g) + 4 H2O(g) 2 CH2Cl2(l) + 2 O2(g) + 4 H2(g) DeltaH = 2 * 16 kJ = 32 kJ
Now, we can add these three equations together to obtain the desired equation:
2 COCl2(g) + 4 H2O(g) + 2 HCl(g) 2 CH2Cl2(l) + 2 O2(g) + 5 H2(g) + Cl2(g) DeltaH = 32 kJ + 167 kJ + 184 kJ = 383 kJ
Therefore, the heat of reaction for the given equation H2O(g) + Cl2(g) 2 HCl(g) + ½ O2(g) is 383 kJ.
Note: Hess's Law states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual steps of the reaction, as long as the equations are multiplied accordingly and their signs are taken into account.