# Question 6a733

Apr 7, 2017

+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 ${\Delta}_{f} H$ of the direct route.
Route B is the $\Sigma {\Delta}_{f} H$ of the reactants
Route C is the $\Sigma {\Delta}_{f} H$ of the products which is the unknown.

${H}_{2}$ + $B {r}_{2}$ $\to$ 2HBr = -72kJ Route A

${H}_{2}$-> 2H =436kJ Route B

$B {r}_{2}$ $\to$ 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.

$\frac{588}{2}$ = +294kJ

Oct 3, 2017

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

#### Explanation:

You get the answer by applying Hess's Law.

You are given three equations:

$\boldsymbol{\text{(1)"color(white)(m) "H"_2"(g)" + "Br"_2"(g)"color(white)(l) → "2HBr(g)" ;color(white)(ll) ΔH =color(white)(ll) "-72 kJ}}$
$\boldsymbol{\text{(2)"color(white)(m) "H"_2"(g)" → "2H(g)";color(white)(mmmmmmm) ΔH = color(white)(l)"436 kJ}}$
$\boldsymbol{\text{(3)"color(white)(m) "Br"_2"(g)" → "2Br(g)";color(white)(mmmmmml) ΔH = color(white)(l)"224 kJ}}$

From these, you must devise the target equation

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

The target equation has $\text{H(g)}$ on the left, so you reverse equation (2) and divide it by 2.

$\boldsymbol{\text{(4)"color(white)(m)"H(g)"→ 1/2"H"_2"(g)";ΔH =color(white)(ll) "-218 kJ}}$

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 $\frac{1}{2} \text{H"_2"(g)}$ on the right, and that is not in the target equation.

You need an equation with $\frac{1}{2} \text{H"_2"(g)}$ on the left.

Divide equation (1) by 2.

$\boldsymbol{\text{(5)"color(white)(m) 1/2"H"_2"(g)" + 1/2"Br"_2"(g)"color(white)(l) → "HBr(g)" ;color(white)(ll) ΔH =color(white)(ll) "-36 kJ}}$

Equation (5) has $\frac{1}{2} \text{Br"_2"(g)}$ on the left, and that is not in the target equation.

You need an equation with $\frac{1}{2} \text{Br"_2"(g)}$ on the right.

Reverse equation (3) and divide it by 2.

$\boldsymbol{\text{(6)"color(white)(m) "Br(g)" → 1/2"Br"_2"(g)"; ΔH = "-112 kJ}}$

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):

$\boldsymbol{\text{(4)"color(white)(m)"H(g)"→color(red)(cancel(color(black)( 1/2"H"_2"(g)")));color(white)(mmmmmmmml)ΔH = "-218 kJ}}$
$\boldsymbol{\text{(5)"color(white)(m) color(red)(cancel(color(black)(1/2"H"_2"(g)"))) + color(red)(cancel(color(black)(1/2"Br"_2"(g)")))color(white)(l) → "HBr(g)" ;color(white)(ll) ΔH =color(white)(l) "-36 kJ}}$
$\underline{\boldsymbol{\text{(6)"color(white)(m) "Br(g)" → color(red)(cancel(color(black)(1/2"Br"_2"(g)")));color(white)(mmmmmmmll) ΔH = "-112 kJ}}}$
$\boldsymbol{\text{(7)" color(white)(m)"H(g)" + "Br(g)" → "HBr(g)"; color(white)(mmmll)Δ_text(rxn)H = "-366 kJ}}$