Question

Thermochemistry: Use the equations to answer the question.?

Answer 1

This question demands an understanding of Hess's Law,

Consider,

`C_2H_4(g) + 3O_2(g) to 2CO_2(g) + 2H_2O(g)`

where `DeltaH_1^° = -1410.1kJ`, and,

`C_4H_8(g) + 6O_2(g) to 4CO_2(g) + 4H_2O(g)`

where `DeltaH_2^° = -2698.3kJ`

and the desired product,

`2C_2H_4 to C_4H_8`

This is, to be sure, the first reaction with double the amounts of reactants, and the second reaction reversed. Hence,

`2*DeltaH_1^° - DeltaH_2^° approx -121.7kJ`

The answer you selected should be right. If it's wrong, I would need to rethink the question.

Answer 2

Warning! Long Answer. B) -121.9 kJ

Explanation:

This is a problem in which you must use Hess' Law, which states that the energy involved in a chemical process is the same whether the process takes place in one or in several steps.

You are given two equations:

`bb"(1)"color(white)(m) "C"_2"H"_4"(g)" + "3O"_2"(g)" → "2CO"_2"(g)" + "2H"_2"O(g)"; color(white)(m)Δ_text(c)H^@ = "-1410.1kJ"`
`bb"(2)"color(white)(m) "C"_4"H"_8"(g)" + "6O"_2"(g)"color(white)(l) → "4CO"_2"(g)" + "4H"_2"O(g)"; color(white)(ll)Δ_text(c)H^@ = "-2698.3 kJ"`

From these, you must devise the target equation

`bb"(3)"color(white)(m)"2C"_2"H"_4"(g)" → "C"_4"H"_8"(g)"; Δ_text(rxn)H^@ = ?`

The target equation has `"2C"_2"H"_4"(g)"` on the left, so you double equation (1).

`bb"(4)"color(white)(m) "2C"_2"H"_4"(g)" + "6O"_2"(g)" → "4CO"_2"(g)" + "4H"_2"O(g)"; Δ_text(rxn)H^@ = "-2820.2 kJ"`

When you double an equation, you double its `ΔH`.

Equation (4) has `"6O"_2"(g)"` on the left, and that is not in the target equation.

You need an equation with `"6O"_2"(g)"` on the right.

Reverse Equation (2).

`bb"(5)"color(white)(m)"4CO"_2"(g)" + "4H"_2"O(g)" → "C"_4"H"_8"(g)" + "6O"_2"(g)" → ; Δ_text(c)H^@ = "+2698.3 kJ"`

When you reverse an equation, you change the sign of its `ΔH`.

Now, you add equations (4) and (5), cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their `ΔH` values.

This gives us the target equation (6):

`bb"(4)"color(white)(m) "2C"_2"H"_4"(g)" + color(red)(cancel(color(black)("6O"_2"(g)"))) → color(red)(cancel(color(black)("4CO"_2"(g)"))) + color(red)(cancel(color(black)("4H"_2"O(g)"))); Δ_text(rxn)H^@ = color(white)(l)"-2820.2 kJ"`
`bb"(5)"ul(color(white)(m)color(red)(cancel(color(black)("4CO"_2"(g)"))) + color(red)(cancel(color(black)("4H"_2"O(g)"))) → "C"_4"H"_8"(g)" + color(red)(cancel(color(black)("6O"_2"(g)"))) ;color(white)(l) Δ_text(c)H^@color(white)(m)= "+2698.3 kJ")`

`bb"(6)"color(white)(m)"2C"_2"H"_4"(g)" → "C"_4"H"_8"(g)"; color(white)(mmmmmmmmmmmml)Δ_text(rxn)H^@ = color(white)(ml)"-121.9 kJ"`

When you add two equations, you add their `ΔH` values.

`Δ_text(rxn)H^@ = "-121.9 kJ"`