Calculate the ΔHrxn of the following reaction using the list of reactions below:
$$2N_{2,(g)}+5O_{2,(g)}rightarrow 2N_2O_{5,(g)}$$ $$H_{2,(g)}+1/2O_{2,(g)}rightarrow H_2O_{(l)},,Delta H_f^circ=-285.8frac{kJ}{mol}$$ $$N_2O_{5,(g)}+H_2O_{(l)}rightarrow 2HNO_{3,(l)},,Delta H_{rxn}=-76.6frac{kJ}{mol}$$ $$N_{2,(g)}+3O_{2,(g)}+H_{2,(g)}rightarrow 2HNO_{3,(l)},,Delta H_{rxn}=-348.2frac{kJ}{mol}$$
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Our goal here is to arrange these three equations so that they add up to the top equation. Every action we do to an equation, we must also do to its enthalpy value.
For our first step, we will flip the second equation and multiply it by 2. This is so we have 2N2O5(g) on the right side. This means we have to multiply the enthalpy value by -2 (the negative is because of the flip). So our new array looks like this: $$H_{2,(g)}+1/2O_{2,(g)}rightarrow H_2O_{(l)},,Delta H_f^circ=-285.8frac{kJ}{mol}$$ $$4HNO_{3,(l)}rightarrow 2N_2O_{5,(g)}+2H_2O_{(l)},,Delta H_{rxn}=(-76.6frac{kJ}{mol}*-2)=153.2frac{kJ}{mol}$$ $$N_{2,(g)}+3O_{2,(g)}+H_{2,(g)}rightarrow 2HNO_{3,(l)},,Delta H_{rxn}=-348.2frac{kJ}{mol}$$
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For our next step, we will flip the top equation and multiply it by 2. This is so the H2O molecules will cancel, since there are none present in the main equation. (If a species is on both the reactant and product side, it will cancel). $$2H_2O_{(l)}rightarrow 2H_{2,(g)}+O_{2,(g)},,Delta H_f^circ=(-285.8frac{kJ}{mol}*-2)=571.6frac{kJ}{mol}$$ $$4HNO_{3,(l)}rightarrow 2N_2O_{5,(g)}+2H_2O_{(l)},,Delta H_{rxn}=153.2frac{kJ}{mol}$$ $$N_{2,(g)}+3O_{2,(g)}+H_{2,(g)}rightarrow 2HNO_{3,(l)},,Delta H_{rxn}=-348.2frac{kJ}{mol}$$
Next, we will multiply the bottom equation by 2. This is for a few reasons: 1) to have 2N2 molecules on the left like the main reaction 2) to cancel out the H2 and HNO3 molecules and 3) to cancel out 1 mol of O2 so we have 5O2 on the left like the main equation. $$2H_2O_{(l)}rightarrow 2H_{2,(g)}+O_{2,(g)},,Delta H_f^circ=571.6frac{kJ}{mol}$$ $$4HNO_{3,(l)}rightarrow 2N_2O_{5,(g)}+2H_2O_{(l)},,Delta H_{rxn}=153.2frac{kJ}{mol}$$ $$2N_{2,(g)}+6O_{2,(g)}+2H_{2,(g)}rightarrow 4HNO_{3,(l)},,Delta H_{rxn}=(-348.2frac{kJ}{mol}*2)=-696.4frac{kJ}{mol}$$
Lastly, we add up the three reaction enthalpies to get the net reaction enthalpy: $$2N_{2,(g)}+5O_{2,(g)}rightarrow 2N_2O_{5,(g)},,Delta H_{rxn}=(571.6frac{kJ}{mol}+153.2frac{kJ}{mol}-696.4frac{kJ}{mol})=28.4frac{kJ}{mol}$$
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