Using the following data, calculate the heat of reaction for the reaction,
1/2 Cl2(g) + 1/2 I2(g) → ICl in kJ/mol:
ΔH° (kJ/mol)
Cl2(g) → 2Cl(g) 242.3
I2(g) → 2I(g) 151.0
ICl(g) → I(g) + Cl(g) 211.3
I2(s) → I2(g) 62.8
a. –211 kJ/mol
b. –14.6 kJ/mol
c. 16.8 kJ/mol
d. 245 kJ/mol
e. 439 kJ/mol
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To find the heat of reaction for the given reaction, we need to calculate the sum of the enthalpy changes for the individual reactions involved.
The given reaction is:
1/2 Cl2(g) + 1/2 I2(g) → ICl
First, we need to convert the enthalpy change for the reaction I2(s) → I2(g) from a solid to a gas.
The reaction I2(s) → I2(g) has an enthalpy change of 62.8 kJ/mol.
Next, we can calculate the enthalpy change for the reaction Cl2(g) → 2Cl(g):
ΔH1 = 2Cl(g) - Cl2(g) = 2 * 0 - 242.3 = -242.3 kJ/mol
The enthalpy change for the reaction I2(g) → 2I(g) is 151.0 kJ/mol.
Finally, the enthalpy change for the reaction ICl(g) → I(g) + Cl(g) is 211.3 kJ/mol.
Now, we can add up the enthalpy changes for the individual reactions to get the enthalpy change for the given reaction:
ΔH = 1/2 ΔH1 + 1/2 ΔH2 + ΔH3 = 1/2 * (-242.3) + 1/2 * 151.0 + 211.3 = -211.0 kJ/mol
Therefore, the heat of reaction for the given reaction, 1/2 Cl2(g) + 1/2 I2(g) → ICl, is -211 kJ/mol, option a.
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