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 (-2755 kJ/mol), and H2 (-286 kJ/mol), calculate H for the reaction.
C4H4(g) + 2 H2(g) C4H8(g)
very confused
What they want you to do here is to use Hess' Law. Write the reactions as combustion reactions.
C4H4 + 5O2 ==> 4CO2 + 2H2O -2341 kJ/mol
2H2 + O2 =>2H2O -286 kJ/MOL= -572 rxn
C4H8 + 6O2 ==> 4CO2 + 4H2O -2755 kJ/mol
========================================
Add the sum of equation 1 and equation 2 to the reverse of equation 3 to give you the reaction you have written (You shouldn't omit the arrow--it makes us guess where the arrow should be). Delta H for the reaction you have written will then be the sum of delta H for equation 1 and equation 2 + delta H for equation 3 (with the sign changed since you changed the direction of the reaction). I've almost worked the entire problem for you.
Post your work if you get stuck.
To calculate the enthalpy change (ΔH) for the given reaction, we can use 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.
First, let's write the balanced chemical equations for the combustion reactions of each compound:
1) C4H4(g) + 5 O2(g) -> 4 CO2(g) + 2 H2O(g) (Equation 1)
2) C4H8(g) + 6 O2(g) -> 4 CO2(g) + 4 H2O(g) (Equation 2)
3) 2 H2(g) + O2(g) -> 2 H2O(g) (Equation 3)
We can see that Equation 3 is the reverse of the water formation step in Equation 1. Therefore, we need to reverse Equation 3 and multiply its enthalpy change by -1.
Now, we can add these equations to get the desired reaction:
C4H4(g) + 2 H2(g) -> C4H8(g)
To calculate the enthalpy change for the reaction, we need to consider the stoichiometric coefficients of the balanced equations.
Multiplying Equation 1 by 2, and Equation 2 by -1 to get the same number of moles of C4H8:
2 C4H4(g) + 10 O2(g) -> 8 CO2(g) + 4 H2O(g) (Equation 4)
- C4H8(g) - 6O2(g) -> - 4CO2(g) - 4H2O(g) (Equation 5)
Now we can add Equations 4 and 5 to obtain the desired reaction:
2 C4H4(g) - C4H8(g) + 6 O2(g) -> 8 CO2(g) + 4 H2O(g) - 4 CO2(g) - 4 H2O(g)
Simplifying:
2 C4H4(g) - C4H8(g) + 6 O2(g) -> 4 CO2(g) + 4 H2O(g)
Now, let's calculate the enthalpy change using the given enthalpies of combustion. We need to consider that the reaction has been rearranged, so the enthalpy change will have the same magnitude but the opposite sign.
ΔH = -[ΔH(C4H4) - ΔH(C4H8) + 6 * ΔH(H2)]
Substituting the given values:
ΔH = -[(-2341 kJ/mol) - (-2755 kJ/mol) + 6 * (-286 kJ/mol)]
Calculating:
ΔH = -[2341 kJ/mol + 2755 kJ/mol - 1716 kJ/mol]
ΔH = -237 kJ/mol
Therefore, the enthalpy change for the reaction C4H4(g) + 2 H2(g) -> C4H8(g) is -237 kJ/mol.
To calculate the enthalpy change, ΔH, for the combustion reaction of C4H4 with H2 to form C4H8, you need to use the Hess's Law equation:
ΔH = Σ(ΔHf products) - Σ(ΔHf reactants)
ΔHf represents the standard enthalpy of formation, which is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states (usually at 25°C and 1 atm).
First, you need to determine the balanced equation for the combustion reaction. Start by balancing the carbon and hydrogen atoms on both sides of the equation:
C4H4(g) + 2 H2(g) → C4H8(g)
Now you can calculate the ΔH for the combustion reaction using the enthalpies of formation. The enthalpy change for the formation of a compound can be found by multiplying the coefficient of the compound by its enthalpy of formation. The enthalpies of formation for C4H4, C4H8, and H2 are given:
ΔHf(C4H4) = -2341 kJ/mol
ΔHf(C4H8) = ? (unknown)
ΔHf(H2) = -286 kJ/mol
Now substitute these values into the equation and solve for ΔH:
ΔH = (1 * ΔHf(C4H8)) - (1 * ΔHf(C4H4) + 2 * ΔHf(H2))
Using the given enthalpies of formation, you can solve for ΔH:
ΔH = (1 * ΔHf(C4H8)) - (1 * -2341 kJ/mol + 2 * -286 kJ/mol)
Simplifying,
ΔH = ΔHf(C4H8) + 4702 kJ/mol + 572 kJ/mol
Combining the constants,
ΔH = ΔHf(C4H8) + 5274 kJ/mol
To get the value of ΔHf(C4H8), you subtract 5274 kJ/mol from both sides of the equation:
ΔH - 5274 kJ/mol = ΔHf(C4H8)
Therefore, the enthalpy change for the reaction is equal to the enthalpy of formation of C4H8 minus 5274 kJ/mol. Make sure to check any values provided in your resources for ΔHf(C4H8).