The enthalpy change of the reaction below (ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
The bond energies data is given as follows:
BE for C≡O = 1072 kJ/mol
BE for Cl-Cl = 242 kJ/mol
BE for C-Cl = 328 kJ/mol
BE for C=O = 766 kJ/mol
The enthalpy change for the reaction is given as :
ΔHr×n = ∑H reactant bond - ∑H product bond
ΔHr×n = ( BE C≡O + BE Cl-Cl) - ( BE C=O + BE 2 × Cl-Cl )
ΔHr×n = ( 1072 + 242 ) - ( 766 + 656 )
ΔHr×n = 1314 - 1422
ΔHr×n = - 108 kJ
Thus, The enthalpy change of the reaction below ( ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
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When the balanced equation for this reaction is:
2Fe + 3H2O → Fe2O3 + 3H2
and according to the vapour pressure formula:
PV= nRT
when we have P is the vapor pressure of H2O= 0.121 atm
and V is the volume of H2O = 4.5 L
and T in Kelvin = 52.5 +273 = 325.5 K
R= 0.08205 atm-L/g mol-K
So we can get n H2O
So, by substitution:
n H2O = PV/RT
= (0.121*4.5)/(0.08205 * 325.5) = 0.02038 gmol
n Fe2O3 = 0.02038 * (1Fe2O3/ 3H2O) = 0.00679 gmol
Note: we get (1FeO3/3H2O) ratio from the balanced equation.
we can get the Mass of Fe2O3 from this formula:
Mass = number of moles * molecular weight
when we have a molecular weight of Fe2O3 = 159.7
= 0.00679 * 159.7 = 1.084 g
∴ 1.084 gm of Fe2O3 will produced
Let's go over the given information. We have the volume, temperature and pressure. From the ideal gas equation, that's 4 out of 5 knowns. So, we actually don't need Pvap of water anymore. Assuming ideal gas, the solution is as follows:
PV=nRT
Solving for n,
n = PV/RT = (753 torr)(1 atm/760 torr)(195 mL)(1 L/1000 mL)/(0.0821 L·atm/mol·K)(25+273 K)
n = 7.897×10⁻³ mol H₂
The molar mass of H₂ is 2 g/mol.
Mass of H₂ = 7.897×10⁻³ mol * 2 g/mol = <em>0.016 g H₂</em>
Answer:
Density (ρ) = 5 kilogram/cubic meter
Explanation:
Steps:
ρ =
m
V
=
10 kilogram
2 cubic meter
= 5 kilogram/cubic meter
