Combustion reactions involve reacting a substance with oxygen. When compounds containing carbon and hydrogen are combusted, carbon dioxide and water are the products. Using the enthalpies of combustion for C4H4 (-2341 kJ/mol), C4H8 (-2766 kJ/mol) and H2 (-286 kJ/mol), calculate delta H for the reaction
C4H4(g) + 2H2(g) --> C4H8(g)
First, we need to calculate the enthalpy of combustion for C4H4 and C4H8 using the given enthalpies of combustion:
C4H4(g) + 5O2(g) --> 4CO2(g) + 2H2O(l)
Delta H = -2341 kJ/mol
C4H8(g) + 7O2(g) --> 4CO2(g) + 4H2O(l)
Delta H = -2766 kJ/mol
Next, we need to balance the equation for the given reaction:
C4H4(g) + 2H2(g) --> C4H8(g)
The reaction is already balanced, so we can proceed to calculate the delta H for the reaction. We can use the enthalpies of combustion for C4H4 and C4H8 to determine the delta H for the reaction:
C4H4(g) + 2H2(g) --> C4H8(g)
Delta H = (-2341 kJ/mol) + (2 * (-286 kJ/mol)) - (-2766 kJ/mol)
Delta H = -2822 kJ/mol
Therefore, the delta H for the reaction is -2822 kJ/mol.
Well, it looks like we have a good old-fashioned combustion reaction on our hands. Now, let's break down the reaction step by step.
First, we need to balance the equation, making sure we have an equal number of atoms on each side:
C4H4(g) + 2H2(g) --> C4H8(g)
Next, we need to figure out the combustion reactions for each of the compounds. The enthalpy of combustion is the energy released when one mole of a substance is completely burned in excess oxygen.
For C4H4, the enthalpy of combustion is -2341 kJ/mol.
For C4H8, the enthalpy of combustion is -2766 kJ/mol.
For H2, the enthalpy of combustion is -286 kJ/mol.
Now, let's calculate the overall enthalpy change (delta H) for the reaction. Since we're going from C4H4 to C4H8, we'll subtract the enthalpy of combustion for C4H4 from the enthalpy of combustion for C4H8:
Delta H = -2766 kJ/mol - (-2341 kJ/mol)
Delta H = -2766 kJ/mol + 2341 kJ/mol
Delta H = -425 kJ/mol
So, the delta H for the reaction C4H4(g) + 2H2(g) --> C4H8(g) is -425 kJ/mol.
To calculate the delta H for the given combustion reaction, we need to use the enthalpies of combustion for each compound involved.
The balanced chemical equation for the reaction is:
C4H4(g) + 2H2(g) → C4H8(g)
The enthalpy change for the reaction can be calculated using the enthalpies of combustion for each compound involved. The enthalpy change for a reaction can be determined using the following equation:
ΔH = Σ(ΔHf products) - Σ(ΔHf reactants)
Where:
ΔHf = standard enthalpy of formation
First, we need to determine the enthalpy of formation for the reactants and products:
ΔHf(C4H4) = -2341 kJ/mol
ΔHf(C4H8) = Unknown
ΔHf(H2) = -286 kJ/mol
Next, we can substitute the values in the equation to calculate the enthalpy change:
ΔH = [ΔHf(C4H8) + 3(ΔHf(H2))] - [ΔHf(C4H4) + 2(ΔHf(H2))]
Since there are 2 moles of H2 in the reactants, we need to multiply the enthalpy of formation of H2 by 2.
ΔH = ΔHf(C4H8) + 3(-286 kJ/mol) - ΔHf(C4H4) - 2(-286 kJ/mol)
Simplifying the equation:
ΔH = ΔHf(C4H8) - ΔHf(C4H4) - 572 kJ/mol
Now, we can substitute the known values:
ΔH = Unknown - (-2341 kJ/mol) - 572 kJ/mol
Simplifying further:
ΔH = Unknown + 2341 kJ/mol - 572 kJ/mol
ΔH = Unknown + 1769 kJ/mol
Therefore, the delta H for the reaction C4H4(g) + 2H2(g) → C4H8(g) is Unknown + 1769 kJ/mol.
To calculate the enthalpy change (ΔH) for a reaction, we need to use the enthalpies of combustion for the reactants and products involved.
The combustion of C4H4 produces 2341 kJ/mol, C4H8 produces -2766 kJ/mol, and H2 produces -286 kJ/mol.
Given the balanced equation: C4H4(g) + 2H2(g) → C4H8(g)
We can calculate the ΔH of this reaction by using the enthalpies of combustion:
ΔH = Σ(H(products)) - Σ(H(reactants))
Let's calculate the enthalpies of the reactants and products:
H(C4H4) = -2341 kJ/mol
H(C4H8) = -2766 kJ/mol
H(H2) = -286 kJ/mol
Now, let's plug these values into the equation:
ΔH = [H(C4H8) + 2 * H(H2)] - H(C4H4)
ΔH = [-2766 kJ/mol + 2 * (-286 kJ/mol)] - (-2341 kJ/mol)
ΔH = -2766 kJ/mol - 572 kJ/mol + 2341 kJ/mol
ΔH = -997 kJ/mol
Therefore, the enthalpy change for the given reaction is -997 kJ/mol.