The following heats of formation are used to find the change in enthalpy for the formation of tin chloride in the following reaction Sn (s) + 2Cl2(g) --> SnCl4(l).Sn(s) + Cl2(g) --> Sn Cl2(l) DH = -186.2 kJSnCl2(s) + Cl2(g) --> SnCl4(l) DH = -325.1 kJWhat is the change in enthalpy?
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To find the change in enthalpy for the formation of tin chloride (SnCl4) in the given reaction, we need to use the heats of formation (ΔHf) for each individual reactant and product.
The change in enthalpy (ΔH) can be calculated using the following equation:
ΔH = Σ(ΔHf products) - Σ(ΔHf reactants)
Let's break down the given reaction and calculate the change in enthalpy step by step:
1. Sn(s) + 2Cl2(g) --> SnCl4(l)
The heat of formation for SnCl4 is not given directly, but we can calculate it by summing up the heats of formation of Sn(s) and Cl2(g).
ΔHf (SnCl4) = ΔHf (Sn(s)) + 4 * ΔHf (Cl2(g))
2. Sn(s) + Cl2(g) --> SnCl2(l)
The heat of formation for SnCl2 is given directly as -186.2 kJ.
3. SnCl2(s) + Cl2(g) --> SnCl4(l)
The heat of formation for SnCl2 is given as -325.1 kJ.
Now, let's substitute the given values into the equation and solve for the change in enthalpy:
ΔH = [ΔHf (SnCl4)] - [ΔHf (Sn(s)) + 4 * ΔHf (Cl2(g))]
ΔH = [-325.1 kJ] - [-186.2 kJ + 4 * ΔHf (Cl2(g))]
At this point, we need the heat of formation of Cl2(g) to proceed further. Since it is not provided in the given information, we won't be able to calculate the change in enthalpy without it.
Therefore, without the heat of formation for Cl2(g), we cannot determine the change in enthalpy for the formation of tin chloride (SnCl4) in the given reaction.