The bombardier beetle uses an explosive discharge as a...

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: $$ \mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{OH})_{2}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ Calculate $\Delta H$ for this reaction from the following data: $\mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}(a q)+\mathrm{H}_{2}(g)$ $$ \begin{aligned} \Delta H &=+177.4 \mathrm{~kJ} \\ \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2}(a q) & \Delta H=-191.2 \mathrm{~kJ} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) & \Delta H=-241.8 \mathrm{~kJ} \\ \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & \Delta H=-43.8 \mathrm{~kJ} \end{aligned} $$

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:
$$
\mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{OH})_{2}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)
$$
Calculate $\Delta H$ for this reaction from the following data:
$\mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}(a q)+\mathrm{H}_{2}(g)$
$$
\begin{aligned}
\Delta H &=+177.4 \mathrm{~kJ} \\
\mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2}(a q) & \Delta H=-191.2 \mathrm{~kJ} \\
\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) & \Delta H=-241.8 \mathrm{~kJ} \\
\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & \Delta H=-43.8 \mathrm{~kJ}
\end{aligned}
$$

Key Concepts

Hess's Law

Hess's Law is the principle that the overall enthalpy change of a reaction is independent of the path between the initial and final states. This allows us to calculate the enthalpy change of a complex reaction by adding, subtracting, or scaling the enthalpy changes of simpler, related reactions.

Reaction Enthalpy

Reaction enthalpy refers to the heat absorbed or released during a chemical reaction under constant pressure. It is a measure of the energy change that accompanies the transformation of reactants into products and indicates whether a reaction is exothermic or endothermic.

Thermochemical Equations

Thermochemical equations are chemical equations that include the enthalpy change associated with the reaction. They enable the use of algebraic manipulation, such as reversing or combining equations, to determine the enthalpy change of a target reaction by relating it to known reactions.

Phase Change Enthalpy

Phase change enthalpy accounts for the energy change that occurs when a substance undergoes a phase transition, such as condensation of a gas to a liquid. In reaction enthalpy calculations, it's important to include these values if the reactants or products are in non-standard states, ensuring an accurate overall energy assessment.

Transcript

00:01 All right.

00:02 We have lots of information to process through in this problem, but the main goal is we want to figure out the delta h of this desired reaction.

00:12 So we need to start manipulating some of our problems in order to, some of our steps in order to get there.

00:18 So the first step, we actually don't need to do anything to because the c6h4 .o .h2 and the c6h4o2 are in the correct places.

00:30 Is they have the correct coefficients.

00:33 So this first one, we're actually not going to do anything to that first equation.

00:36 Now, this second one has the h2o aqueous in it, and it's on the wrong side.

00:42 So what we're going to do is we're going to flip this reaction.

00:46 So we're going to flip it around.

00:48 So that way we have h2o2 aqueous.

00:53 On the correct side is a reactant instead of a product, making h2 gas plus o2 gas, and then since we flipped the equation, we're going to flip the sign of our delta h.

01:10 So this equation is now finished.

01:13 I'm going to erase it so we don't get confused.

01:21 Next we have this h2o liquid.

01:25 So i'm going to use this last equation here because that's the one that has it.

01:29 This middle equation will use just to make sure things are canceling out correctly.

01:33 So there's we need, it's on the correct side, but we need two of them.

01:39 So we're going to multiply everything by two.

01:41 So we'll have 2h2o gas, making 2h2o liquid...

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