From the enthalpies of reaction
2C(s)+O2(g)�¨2CO(g)ƒ¢H=−221.0kJ
2C(s)+O2(g)+4H2(g)�¨2CH3OH(g)ƒ¢H=−402.4kJ
calculate ĢH for the reaction
CO(g)+2H2(g)�¨CH3OH(g)
To calculate the enthalpy change (ΔH) for the reaction CO(g) + 2H2(g) → CH3OH(g), we can use the concept of Hess's Law.
Hess's Law states that the total enthalpy change for a reaction is the same whether it occurs in one step or through a series of intermediate steps.
In this case, we need to manipulate the given reactions so that the desired reaction is obtained. We can cancel out the common compounds from both reactions to form the desired equation.
1. Reverse the first reaction:
2CO(g) → 2C(s) + O2(g)
ΔH = +221.0 kJ
2. Multiply the second reaction by 2:
4C(s) + 2O2(g) + 8H2(g) → 4CH3OH(g)
ΔH = -804.8 kJ
3. Multiply the first reaction by 2:
4CO(g) → 4C(s) + 2O2(g)
ΔH = +442.0 kJ
4. Now, add the manipulated reactions together:
4CO(g) + 8H2(g) → 4CH3OH(g)
-804.8 kJ + 442.0 kJ = -362.8 kJ
Therefore, the enthalpy change (ΔH) for the reaction CO(g) + 2H2(g) → CH3OH(g) is -362.8 kJ.
To calculate the enthalpy change (ΔH) for the given reaction, we can use Hess's Law. According to Hess's Law, the enthalpy change for a reaction can be determined by adding or subtracting the enthalpies of other reactions that can be combined to yield the desired reaction.
Looking at the given reactions, we need to manipulate them to obtain the desired reaction: CO(g) + 2H2(g) --> CH3OH(g)
Firstly, we need to reverse the direction of the first reaction, so it becomes CO(g) --> C(s) + 0.5O2(g). By doing this, the sign of the enthalpy change also changes, so it becomes ΔH = +221.0 kJ.
Next, we need to double the second reaction to obtain an equal number of moles of CO(g). Therefore, for the second reaction, it becomes 2C(s) + 2O2(g) + 4H2(g) --> 2CH3OH(g), with ΔH = -402.4 kJ.
Now, we can cancel out the reactant and product that are common to both the reversed first reaction and the second reaction, which is C(s), O2(g), and 2CH3OH(g). This leaves us with:
C(s) + 0.5O2(g) + 2H2(g) --> CH3OH(g)
To find ΔH for the desired reaction, we add the ΔH values of the manipulated reactions. In this case, it would be:
ΔH = (+221.0 kJ) + (-402.4 kJ) = -181.4 kJ
Therefore, the enthalpy change for the reaction CO(g) + 2H2(g) --> CH3OH(g) is -181.4 kJ.
Reverse equation 1 and add to equation 2. Divide the final equation by 2.
Add the dH values; when an equation is reversed, change the sign of dH.