Question #6a733

2 Answers
Apr 7, 2017

Answer:

+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

Oct 3, 2017

Answer:

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"#