The bombardier beetle uses an explosive discharge as a defensive measure. The chemical reaction involved is th?

The bombardier beetle uses an explosive discharge as a defensive measure. The chemical reaction involved is the oxidation of hydroquinone by hydrogen peroxide to produce quinone and water.

C6H4(OH)2(aq) + H2O2(aq) C6H4O2(aq) + 2 H2O(l)

Calculate ΔH for the above reaction from the following data.

C6H4(OH)2(aq) C6H4O2(aq) + H2(g) ΔH = +177.4 kJ

H2(g) + O2(g) H2O2(aq) ΔH = -191.2 kJ

H2(g) + 1/2 O2(g) H2O(g) ΔH = -241.8 kJ

H2O(g) H2O(l) ΔH = -43.8 kJ

can u please help me! and show me a step by step process and answer!?

4 Answers

  • the idea is you multiply the equations and or flip them around to add up to the equation you need. And when you flip an equation around, the sign of ΔH changes.. like this

    you want

    C6H4(OH)2(aq) + H2O2(aq) -- > C6H4O2(aq) + 2 H2O(l)

    and you have these starting equations:

    (1).. C6H4(OH)2(aq) ---> C6H4O2(aq) + H2(g)... . ΔH = +177.4 kJ

    (2).. H2(g) + O2(g) ..---> H2O2(aq)... ... ... .. .. .. ΔH = -191.2 kJ

    (3).. H2(g) + 1/2 O2(g) --> H2O(g)... ... .. .. .. .. . .ΔH = -241.8 kJ

    (4).. H2O(g) ---> H2O(l)... ... .. .. .. . .. .. . ... . . .. ΔH = -43.8 kJ

    *******

    equation (1) has C6H4(OH)2(aq) and C6H4O2(aq) in the right positions and the right quantities. So we'll copy it exactly

    (5).. C6H4(OH)2(aq) ---> C6H4O2(aq) + H2(g)... . ΔH = +177.4 kJ

    *******

    now we need to add H2O2(aq) to the left.. and in equation(2).. it's on the right.. so we need to flip it around and change the sign of ΔH

    (5).. C6H4(OH)2(aq) ---> C6H4O2(aq) + H2(g)... . ΔH = +177.4 kJ

    (6).. H2O2(aq) ---> H2(g) + O2(g) ... ... ... .. .. .. ΔH = +191.2 kJ

    *******

    now we need to get rid of the 2 H2(g) and 1 O2 on the right so we'll multiply equation (3) by 2 and add it to the pile. And we have to multiply ΔH by 2 also.

    (5).. C6H4(OH)2(aq) ---> C6H4O2(aq) + H2(g)... . ΔH = +177.4 kJ

    (6).. H2O2(aq) ---> H2(g) + O2(g) ... .... ... .. .. .. ΔH = +191.2 kJ

    (7).. 2 H2(g) + 1 O2(g) --> 2 H2O(g).... .. .. .. .. . .ΔH = -483.6 kJ

    *******

    and now we need to convert 2 H2O(g) to 2 H2O(l).. so we multiply equation (4) by 2 and add it to the pile

    (5).. C6H4(OH)2(aq) ---> C6H4O2(aq) + H2(g)... . ΔH = +177.4 kJ

    (6).. H2O2(aq) ---> H2(g) + O2(g) ... .... ... .. .. .. ΔH = +191.2 kJ

    (7).. 2 H2(g) + 1 O2(g) --> 2 H2O(g).... .. .. .. .. . .ΔH = -483.6 kJ

    (8).. 2 H2O(g) ---> 2 H2O(l)... ... .. .. .. . .. .. . . .. ΔH = -87.6 kJ

    *******

    now when you add them you get

    C6H4(OH)2(aq) + H2O2(aq) -- > C6H4O2(aq) + 2 H2O(l)... ΔH = 202.6 kJ

    *******

    capice?

  • Uses Of Hydroquinone

  • Answer should be negative.

  • i have no idea

Hottest videos

Leave a Reply

Your email address will not be published. Required fields are marked *

Related Posts