Question

Question #6a733

Answer 1

+294kJ

Explanation:

We will use Hess' Law to work out `Delta_fH^`
(Normally an enthalpy cycle would be drawn but I do not know how to do that on a laptop so here is an alternative)

Hess' Law states that there is an indirect route along with the direct route in a reaction. We can use this indirect route to find the enthalpy change of the direct route.

Route A is the `Delta_fH` of the direct route.
Route B is the `SigmaDelta_fH` of the reactants
Route C is the `SigmaDelta_fH` of the products which is the unknown.

`H_2` + `Br_2` `->` 2HBr = -72kJ Route A

`H_2`-># 2H =436kJ Route B

`Br_2` `->` 2Br = 224kJ Route B

We need to find route C which is A+B

224+436= 660kJ

660 - 72 = 588kJ

588kJ is for 2HBr but we only want HBr so we divide 588 by 2.

`588/2` = +294kJ

Answer 2

Warning! Long Answer. `Δ_text(rxn)H = "-366 kJ"`

Explanation:

You get the answer by applying Hess's Law.

You are given three equations:

`bb"(1)"color(white)(m) "H"_2"(g)" + "Br"_2"(g)"color(white)(l) → "2HBr(g)" ;color(white)(ll) ΔH =color(white)(ll) "-72 kJ"`
`bb"(2)"color(white)(m) "H"_2"(g)" → "2H(g)";color(white)(mmmmmmm) ΔH = color(white)(l)"436 kJ"`
`bb"(3)"color(white)(m) "Br"_2"(g)" → "2Br(g)";color(white)(mmmmmml) ΔH = color(white)(l)"224 kJ"`

From these, you must devise the target equation

`"H(g)" + "Br(g)" → "HBr(g)"; Δ_text(rxn)H = ?`

The target equation has `"H(g)"` on the left, so you reverse equation (2) and divide it by 2.

`bb"(4)"color(white)(m)"H(g)"→ 1/2"H"_2"(g)";ΔH =color(white)(ll) "-218 kJ"`

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

When you divide an equation by 2, you halve its `ΔH`.

Equation (4) has `1/2"H"_2"(g)"` on the right, and that is not in the target equation.

You need an equation with `1/2"H"_2"(g)"` on the left.

Divide equation (1) by 2.

`bb"(5)"color(white)(m) 1/2"H"_2"(g)" + 1/2"Br"_2"(g)"color(white)(l) → "HBr(g)" ;color(white)(ll) ΔH =color(white)(ll) "-36 kJ"`

Equation (5) has `1/2"Br"_2"(g)"` on the left, and that is not in the target equation.

You need an equation with `1/2"Br"_2"(g)""` on the right.

Reverse equation (3) and divide it by 2.

`bb"(6)"color(white)(m) "Br(g)" → 1/2"Br"_2"(g)"; ΔH = "-112 kJ"`

Now, you add equations (4), (5) and (6), 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 (7):

`bb"(4)"color(white)(m)"H(g)"→color(red)(cancel(color(black)( 1/2"H"_2"(g)")));color(white)(mmmmmmmml)ΔH = "-218 kJ"`
`bb"(5)"color(white)(m) color(red)(cancel(color(black)(1/2"H"_2"(g)"))) + color(red)(cancel(color(black)(1/2"Br"_2"(g)")))color(white)(l) → "HBr(g)" ;color(white)(ll) ΔH =color(white)(l) "-36 kJ"`
`ul(bb"(6)"color(white)(m) "Br(g)" → color(red)(cancel(color(black)(1/2"Br"_2"(g)")));color(white)(mmmmmmmll) ΔH = "-112 kJ")`
`bb"(7)" color(white)(m)"H(g)" + "Br(g)" → "HBr(g)"; color(white)(mmmll)Δ_text(rxn)H = "-366 kJ"`