Given the standard enthalpy changes for the following two reactions:
(1) Ni(s) + Cl2(g) -> NiCl2(s)...... ΔH° = -305.3 kJ
(2) Fe(s) + Cl2(g) -> FeCl2(s)......ΔH° = -341.8 kJ
what is the standard enthalpy change for the reaction:
(3) Ni(s) + FeCl2(s)NiCl2(s) + Fe(s)......ΔH° = ______kJ
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To find the standard enthalpy change for reaction (3), we can use the fact that enthalpy change is a state function. This means that the overall enthalpy change for a reaction can be calculated by summing the enthalpy changes of the individual steps.
In this case, we can rewrite reaction (3) as the sum of reactions (1) and (2) with the appropriate coefficients:
(3) Ni(s) + FeCl2(s) -> NiCl2(s) + Fe(s)
can be written as:
(4) Ni(s) + Cl2(g) -> NiCl2(s) (reverse of reaction 1)
(5) Fe(s) + Cl2(g) -> FeCl2(s) (reverse of reaction 2)
Adding reactions (4) and (5) yields reaction (3):
(4) + (5) : Ni(s) + Cl2(g) + Fe(s) + Cl2(g) -> NiCl2(s) + FeCl2(s)
Now, we can sum the enthalpy changes for reactions (4) and (5) to find the overall enthalpy change for reaction (3):
(ΔH°)reaction (3) = (ΔH°)reaction (4) + (ΔH°)reaction (5)
(ΔH°)reaction (3) = -305.3 kJ + (-341.8 kJ)
(ΔH°)reaction (3) = -647.1 kJ
Therefore, the standard enthalpy change for reaction (3) is -647.1 kJ.
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