The combustion of 1 mole of C2H5OH (Ethanol) in a bomb calorimeter produces 1364 kJ/mol at 25 degree Calcium. What is the value of enthalpy chamge for the following combustion reaction?
C2H5OH(l) + 3O2(g) => 2CO2(g) + 3H2O(l)
To find the value of the enthalpy change (ΔH) for the given combustion reaction, you need to apply the Hess's Law, which states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of its individual steps.
In this case, we'll consider the following steps:
1) Ethanol (C2H5OH) combustion:
C2H5OH(l) + 3O2(g) => 2CO2(g) + 3H2O(l) ... (1)
2) Combustion of carbon (C):
C(s) + O2(g) => CO2(g) ... (2)
3) Combustion of hydrogen (H2):
H2(g) + 1/2O2(g) => H2O(l) ... (3)
Since the given heat of combustion for ethanol is 1364 kJ/mol, it corresponds to the enthalpy change for equation (1).
Now, we need to use equations (2) and (3) to re-write equation (1) in terms of these equations. By doing so, we can use the given enthalpy change for ethanol and the enthalpy changes for equations (2) and (3) to find the enthalpy change for the entire reaction.
First, we need to match the number of moles of CO2 and H2O from equation (1) to the number of moles in equations (2) and (3). To do this, we multiply equation (2) by 2 and equation (3) by 3:
2 x (C(s) + O2(g) => CO2(g)) => 2CO2(g) ... (4)
3 x (H2(g) + 1/2O2(g) => H2O(l)) => 3H2O(l) ... (5)
Now, we can combine equations (4) and (5) to obtain the overall equation:
2 x (C(s) + O2(g) => CO2(g) + H2(g) + 1/2O2(g) => H2O(l)) => 2CO2(g) + 3H2O(l) ... (6)
Comparing equation (6) with equation (1), we can see that the enthalpy change for equation (6) is equal to the sum of enthalpy changes for equations (2) and (3).
Therefore, we can write the equation:
ΔH_ethanol + ΔH_C + ΔH_H = ΔH_overall
Substituting the given value of ΔH_ethanol = 1364 kJ/mol, we need to find the values of ΔH_C and ΔH_H. The values of ΔH_C and ΔH_H can be found using standard enthalpy changes, which are:
ΔH_C = -394 kJ/mol (standard enthalpy change for the combustion of carbon)
ΔH_H = -286 kJ/mol (standard enthalpy change for the combustion of hydrogen)
Now, we can calculate the overall enthalpy change:
1364 kJ/mol + (-394 kJ/mol) + (-286 kJ/mol) = ΔH_overall
Simplifying the expression, we get:
ΔH_overall = 1364 kJ/mol - 394 kJ/mol - 286 kJ/mol
ΔH_overall = 684 kJ/mol
Therefore, the value of the enthalpy change for the given combustion reaction is 684 kJ/mol.
To find the enthalpy change for the given combustion reaction, we can use the concept of Hess's Law, which states that the enthalpy change for a reaction is the same regardless of the route taken between the reactants and products.
We can break down the given reaction into a series of known reactions for which enthalpy changes are provided:
1. C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)
2. 3H2O(g) → 3H2O(l)
Now, we can use the enthalpy changes provided and apply them to these reactions to calculate the enthalpy change for the given combustion reaction.
Given:
Enthalpy change for the combustion of 1 mole of C2H5OH: -1364 kJ/mol
Enthalpy change for the vaporization of 1 mole of water: +40.7 kJ/mol
Step 1: Balancing the equations
We start by balancing the equations to match the given reaction:
1. C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)
2. 3H2O(g) → 3H2O(l)
Step 2: Applying enthalpy changes
Since the enthalpy change is the same regardless of the route taken, we can reverse equation 2 and multiply it by -1 to calculate the enthalpy change for the conversion of water from gaseous to liquid form:
3H2O(l) → 3H2O(g)
Enthalpy change = -(-40.7 kJ/mol) = +40.7 kJ/mol
Now, to find the enthalpy change for the given combustion reaction, we add up the enthalpy changes of the individual reactions:
Enthalpy change = (Enthalpy change of reaction 1) + (Enthalpy change of reaction 2)
Enthalpy change = -1364 kJ/mol + 40.7 kJ/mol
Enthalpy change = -1323.3 kJ/mol
Therefore, the value of enthalpy change for the given combustion reaction C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) is -1323.3 kJ/mol.