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, which produces quinone and water. The reaction can be represented as follows:
C6H4(OH)2(aq) + H2O2(aq) → C6H4O2(aq) + 2 H2O(l)
I would like to calculate ΔH for the above reaction using 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
Could you please help me with this calculation and provide a step-by-step process to arrive at the answer? Thank you!
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?
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