Given these reactions :
X(s) + 1/2O2(g) --> XO(s) (Delta)H= -707.7 kj
XCO3(s) --> XO(s)+CO2(g) (Delta)H= +381.9kj
What is (delta)Hrxm for this reaction?
X(s)+1/2 O2(g)+CO2(g) ---> XCO3(s)
(Delta)H= ?
Well, well, let's see what we have here! It seems like a good old chemical reaction. Now, to find the ΔH for this reaction, we can simply add up the ΔH values of the individual reactions.
First, we have X(s) + 1/2 O2(g) → XO(s) with a ΔH of -707.7 kJ.
Next, we have XCO3(s) → XO(s) + CO2(g) with a ΔH of +381.9 kJ.
Now, to find the overall ΔH, let's add these two values together:
-707.7 kJ + (+381.9 kJ) = -325.8 kJ
So, my dear friend, the ΔH for the reaction X(s) + 1/2 O2(g) + CO2(g) → XCO3(s) is -325.8 kJ. It seems like this reaction releases some heat. Stay cool, my friend!
To find ΔHrxn for the reaction:
X(s) + 1/2O2(g) + CO2(g) → XCO3(s)
You can use Hess's Law, which states that the enthalpy change of a reaction is independent of the pathway taken and depends only on the initial and final states of the system.
In this case, you need to manipulate the given reactions by multiplying or reversing them to obtain the desired overall reaction:
1) Reverse the second reaction:
XO(s) + CO2(g) → XCO3(s)
ΔH = -381.9 kJ
2) Multiply the first reaction by 2:
2X(s) + O2(g) → 2XO(s)
ΔH = -2(-707.7 kJ) = +1415.4 kJ
Now, add the manipulated reactions together to obtain the overall reaction:
2X(s) + O2(g) + CO2(g) → 2XO(s) + CO2(g) + XCO3(s)
The ΔHrxn for the overall reaction is the sum of the ΔH values for the manipulated reactions:
ΔHrxn = (+1415.4 kJ) + (-381.9 kJ) = +1033.5 kJ
To find the ΔHrxm (enthalpy change of the reaction) for the given chemical equation, you can use Hess's Law, which states that if a reaction can be written as the sum of two or more reactions, then the ΔHrxm of the overall reaction is equal to the sum of the ΔHrxm of the individual reactions.
Here are the steps to find ΔHrxm for the given reaction:
Step 1: Write the given reactions:
1. X(s) + 1/2 O2(g) → XO(s) ΔH = -707.7 kJ
2. XCO3(s) → XO(s) + CO2(g) ΔH = +381.9 kJ
Step 2: Rearrange the given reactions to match the desired equation:
1. XO(s) → X(s) + 1/2 O2(g) ΔH = +707.7 kJ (reversed the first equation)
2. XO(s) + CO2(g) → XCO3(s) ΔH = -381.9 kJ (reversed and changed the sign of the second equation)
Step 3: Multiply each reaction to ensure that the stoichiometric coefficients align with the desired equation:
1. 2XO(s) → 2X(s) + O2(g) ΔH = 2*(-707.7) kJ = -1415.4 kJ (multiplied the first equation by 2)
2. XCO3(s) → XO(s) + CO2(g) ΔH = +381.9 kJ
Step 4: Add the two reactions together to get the desired equation:
2XO(s) + XCO3(s) → 2X(s) + O2(g) + CO2(g) ΔHrxm = (-1415.4 kJ) + (+381.9 kJ)
ΔHrxm = -1033.5 kJ
Therefore, the ΔHrxm for the reaction X(s) + 1/2 O2(g) + CO2(g) → XCO3(s) is -1033.5 kJ.