Please show me how to work!
Ethane, C2H6(g) can be made by reaction of hydrogen gas with acetylene, C2H2(g). the standard enthalpies of formation of ethane and acetylene are -84.68 and +226.73 kJ mol-1, respectively. what is the reaction enthalpy for the production of one mole of ethane? Write the balanced chemical equation for the reaction first.
C2H2 + 2H2 ==> CH3CH3
dHrxn = (n*dHf products) - (n*dH reactants).
dHrxn = (n*dHf products) - (n*ch reactants)
The balanced chemical equation for the reaction can be written as:
2C2H2(g) + 5H2(g) -> C2H6(g)
To calculate the reaction enthalpy, you can use the standard enthalpies of formation. The enthalpy change for the reaction is equal to the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants.
The given standard enthalpies of formation are:
ΔHf° (ethane) = -84.68 kJ mol-1
ΔHf° (acetylene) = +226.73 kJ mol-1
Using the equation, the enthalpy change for the reaction is:
ΔHrxn = (2 x ΔHf° (ethane)) - (2 x ΔHf° (acetylene)) + (5 x 0)
ΔHrxn = (2 x -84.68 kJ mol-1) - (2 x 226.73 kJ mol-1) + (5 x 0)
ΔHrxn = -169.36 kJ mol-1 - (-453.46 kJ mol-1)
ΔHrxn = -169.36 kJ mol-1 + 453.46 kJ mol-1
ΔHrxn = 284.1 kJ mol-1
Therefore, the reaction enthalpy for the production of one mole of ethane is 284.1 kJ mol-1.
To determine the reaction enthalpy for the production of one mole of ethane, we first need to write the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction can be written as follows:
2 C2H2(g) + 5 H2(g) -> 2 C2H6(g)
Now, let's calculate the reaction enthalpy using the enthalpies of formation. The reaction enthalpy is the difference in the sum of the enthalpies of formation of the products and reactants:
ΔH = ∑(ΔHf products) - ∑(ΔHf reactants)
Given the standard enthalpies of formation of ethane (-84.68 kJ mol-1) and acetylene (+226.73 kJ mol-1), we can substitute these values into the equation.
ΔH = (2 × ΔHf C2H6) - (2 × ΔHf C2H2 + 5 × ΔHf H2)
ΔH = (2 × -84.68 kJ mol-1) - (2 × +226.73 kJ mol-1 + 5 × 0 kJ mol-1)
ΔH = -169.36 kJ mol-1 - 453.46 kJ mol-1
ΔH = -622.82 kJ mol-1
Therefore, the reaction enthalpy for the production of one mole of ethane is -622.82 kJ mol-1.