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lukranit [14]
3 years ago
6

HQ5.49 Use standard enthalpies of formation to calculate the standard enthalpy of reaction for the following reaction: 2 H 2 ​ S

(g) + 3 O 2 ​ (g) → 2 H 2 ​ O(l) + 2 SO 2 ​ (g) ΔH ∘ rxn ​
Chemistry
1 answer:
weqwewe [10]3 years ago
5 0
Answer is: <span>the standard enthalpy of reaction is</span> -1124,2 kJ/mol.
ΔHf(H₂S) = -20.5 kJ/mol
ΔHf(O₂)=0 kJ/mol.
ΔHf(H₂O) = -285.8 kJ/mol.
ΔHf(SO₂) = -296.8 kJ/mol.
ΔHrxn = ∑ΔHf (products of reaction) - ∑ΔHf (reactants).<span>
ΔHrxn - enthalpy change of chemical reaction.
<span>ΔHf - enthalpy of formation of reactants or products.
</span></span>ΔHrxn = (2 · (-285,8) + 2 · (-296,8)) - (2 · (-20,5)) · kJ/mol.
ΔHrxn = -1124,2 kJ/mol.

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A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio
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Answer:

C_7H_6O_2

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC

And the moles of hydrogen due to the produced 1.50 grams of water:

n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}  =0.166molH

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO

Thus, the subscripts in the empirical formula are:

C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol

Which are equal to the moles of the acid:

n_{acid}=0.0279mol

And the molar mass:

MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

C_7H_6O_2

Best regards!

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