The chemical reaction representing production of water gas is as follows:
C(s)+H2O(l)=CO(g)+H2(g)
calculate the enthalpy change in the production of 200L(at 500mmHg and 65degree celcius) of hydrogen by this reaction.
I think temperature will be a factor. Ignoring that
figure the moles of H2, and the moles of CO2, and the moles of H2O(l).
ChangeEnthalpy=HformationCO*molesCO -HeatformationH2O*molesH20
Now the problem. Heat of Formation is given in tables at some standard temperature, usually 273K. I hate to tell you to ignore that, as it is significant.
so would i find the moles using the ideal gas law? PV=nRT
i didn't understand the heat of formation? could you explain a little more? thanks
To calculate the enthalpy change (ΔH) in the production of 200 L of hydrogen using the given reaction, you need to use Hess's Law and the enthalpy values of the individual reactions involved.
Hess's Law states that if a reaction can be expressed as the sum of two or more reactions, then the enthalpy change for the overall reaction is the sum of the enthalpy changes of the individual reactions.
In this case, we need to consider two separate reactions:
1. C(s) + O2(g) → CO2(g) ΔH₁ = -393.5 kJ/mol
This reaction represents the production of carbon dioxide.
2. CO(g) + H2(g) → CO2(g) + H2O(l) ΔH₂ = -41.2 kJ/mol
This reaction represents the production of water gas (carbon monoxide and hydrogen).
Given the balanced equation:
C(s) + H2O(l) → CO(g) + H2(g)
We can see that the formation of CO(g) from CO2(g) (reaction 1) is the reverse of the desired reaction, and the formation of H2(g) from H2O(l) (reaction 2) is the same as the desired reaction.
To calculate the enthalpy change for the desired reaction, we need to reverse the equation for reaction 1 and sum it with reaction 2:
CO2(g) → C(s) + O2(g) ΔH₁' = +393.5 kJ/mol (Note the opposite sign from ΔH₁)
CO2(g) + H2(g) → CO(g) + H2O(l) ΔH₂ = -41.2 kJ/mol
Now, we can sum these two reactions to get the desired reaction:
C(s) + H2O(l) + O2(g) → CO(g) + H2(g) + CO2(g) + H2O(l)
Notice that water (H2O) appears on both sides and can be canceled out:
C(s) + O2(g) → CO(g) ΔH₁' = +393.5 kJ/mol
H2(g) → H2(g) (No energy change, as it is unchanged)
CO2(g) + H2(g) → CO(g) + H2O(l) ΔH₂ = -41.2 kJ/mol
Now, we can sum the enthalpy changes:
ΔH_total = ΔH₁' + ΔH₂
ΔH_total = +393.5 kJ/mol - 41.2 kJ/mol
To calculate the enthalpy change in the production of 200 L of hydrogen, we need to convert the given volume to moles. The molar volume of any ideal gas at standard temperature and pressure (STP) is 22.4 L/mol. However, since we have different conditions (500 mmHg and 65 degrees Celsius), we need to adjust the volume using the ideal gas law.
Using the ideal gas law equation: PV = nRT
Rearranging the equation as: n = PV / RT
n = (500 mmHg * 200 L) / (0.0821 * (65 + 273.15) K * mol⁻¹)
where P is pressure, V is volume, R is the ideal gas constant, T is temperature, and n is the number of moles.
Calculate n to get the moles of hydrogen gas produced.
Finally, multiply the moles of hydrogen gas by the enthalpy change (ΔH_total) to get the enthalpy change for the production of 200 L of hydrogen gas.