How do I use Hess's law to calculate enthalpy change? (when given two reactions)
(1)Pb(s) + Cl2(g) --> PbCl2(s) Delta H(1)= -359.4 kJ
(2)Ni(s) + Cl2(g) --> NiCl2(s) Delta H(2) = -305.3 kJ
This is the third reaction:
(3) PbCl2(s) + Ni(s) --> Pb(s) + NiCl2(s) Delta H(3) = ? kJ
I'm confused is it -664.7 or -54.1 or am I wrong? :(
Do this.
Reverse equation 1 and add to equation 2. Do and check to see that this adds to equation 3 which is what you want. Since you reversed equation 1 you want to change sign on dH for that equation and add to dH for equation 2 to obtain the dH for the final equation (eqn 3).
To use Hess's law to calculate the enthalpy change of the third reaction, you need to manipulate the given reactions in such a way that when added, they cancel out all the substances except the desired reaction.
Step 1: Manipulate the given reactions:
Reverse reaction (2) to obtain:
(2') NiCl2(s) --> Ni(s) + Cl2(g) ΔH(2') = +305.3 kJ
Step 2: Multiply the reactions as necessary:
Multiply reaction (1) by 1 and reaction (2') by 1 to make the stoichiometric coefficients match with reaction (3):
(1) Pb(s) + Cl2(g) --> PbCl2(s) ΔH(1) = -359.4 kJ
(2') NiCl2(s) --> Ni(s) + Cl2(g) ΔH(2') = +305.3 kJ
Step 3: Add the manipulated and multiplied reactions to obtain reaction (3):
(1) Pb(s) + Cl2(g) --> PbCl2(s) ΔH(1) = -359.4 kJ
(2') NiCl2(s) --> Ni(s) + Cl2(g) ΔH(2') = +305.3 kJ
-----------------------------------------
(3) PbCl2(s) + Ni(s) --> Pb(s) + NiCl2(s)
Step 4: Add the enthalpy changes:
ΔH(3) = ΔH(1) + ΔH(2')
ΔH(3) = -359.4 kJ + 305.3 kJ
ΔH(3) = -54.1 kJ
Therefore, the enthalpy change for the third reaction is -54.1 kJ.
To use Hess's law to calculate the enthalpy change of the third reaction, you can apply the following steps:
Step 1: Start by writing the balanced equations for reaction (1), reaction (2), and reaction (3).
(1)Pb(s) + Cl2(g) → PbCl2(s) ΔH(1) = -359.4 kJ
(2)Ni(s) + Cl2(g) → NiCl2(s) ΔH(2) = -305.3 kJ
(3) PbCl2(s) + Ni(s) → Pb(s) + NiCl2(s) ΔH(3) = ?
Step 2: Recognize that you can manipulate the given equations to match the desired equation (reaction 3). For example, you can multiply reaction (1) by 1 and reaction (2) by 1 and combine them to give you the desired equation.
Pb(s) + Cl2(g) → PbCl2(s) ΔH(1) = -359.4 kJ
Ni(s) + Cl2(g) → NiCl2(s) ΔH(2) = -305.3 kJ
PbCl2(s) + Ni(s) - Pb(s) + NiCl2(s) ΔH(3) = ?
Step 3: Now, you can add the desired reaction (3) from the manipulated reactions (1 and 2). To do this, you need to add the enthalpy changes for reactions (1) and (2) to obtain the enthalpy change for reaction (3).
ΔH(3) = ΔH(1) + ΔH(2)
ΔH(3) = (-359.4 kJ) + (-305.3 kJ)
ΔH(3) = -664.7 kJ
Therefore, the enthalpy change for reaction (3) is -664.7 kJ.
So, you are correct in your initial calculation. The enthalpy change for reaction (3) is indeed -664.7 kJ, not -54.1 kJ.