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Lilit [14]
2 years ago
13

Consider the reaction H2(g) + Cl2(g) → 2HCl(g)ΔH = −184.6 kJ / mol If 2.00 moles of H2 react with 2.00 moles of Cl2 to form HCl,

what is ΔU (in kJ) for this reaction at 1.0 atm and 25°C? Assume the reaction goes to completion.
Chemistry
1 answer:
zalisa [80]2 years ago
4 0

Answer:

ΔU=-369.2 kJ/mol.

Explanation:

We start from the equation:

Δ(H)=ΔU+Δ(PV), which is an extension of the well known relation: H=U+PV.

If Δ(PV) were calculated by ideal gas law,

PV=nRT

Δ(PV)=RTΔn.

Where Δn is the change of moles due to the reaction; but, this reaction does not give a moles change (Four moles of HCl produced from 4 moles of reactants), so Δ(PV)=0.

So, for this case, ΔH=ΔU.

The enthalpy of reaction given is for one mole of reactant, so the enthalpy of reaction for the reaction of interest must be multiplied by two:

2 reactant moles*\frac{-184.6kJ}{mol}

ΔU=-369.2 kJ/mol.

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Answer:

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2 years ago
You mix 500.0 mL of 0.250 M iron(III) chloride solution with 425.0 mL of 0.350 M barium chloride solution. Assuming the volumes
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Answer:

M=0.727M

Explanation:

Hello,

In this case, since iron (III) chloride (FeCl3) and barium chloride (BaCl2) are both chloride-containing compounds, we can compute the moles of chloride from each salt, considering the concentration and volume of the given solutions, and using the mole ratio that is 1:3 and 1:2 for the compound to chlorine:

n_{Cl^-}=0.50L*0.250\frac{molFeCl_3}{L}*\frac{3molCl^-}{1molFeCl_3}=0.375molCl^-  \\\\n_{Cl^-}=0.425L*0.350\frac{molBaCl_2}{L}*\frac{2molCl^-}{1molBaCl_2}=0.2975molCl^-

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