Calculate the change in enthalpy for the following reaction, using the bond enthalpy data provided below. (Note: you may round t
1 answer:
Answer:
-514 kJ/mol
Explanation:
The bond enthalpy which is also known as bond energy can be defined as the amount of energy needed to split one mole of the stated bond. The change in enthalpy of a given reaction can be estimated by subtracting the sum of the bond energies of the reactants from the sum of the bond energies of the products.
For the given chemical reaction, the change in enthalpy of the reaction is:
Δ [2(409) + 4(388) + 3(496) - 4(630) - 4(463)] kJ/mol = 818 + 1552 + 1488 - 2520 - 1852 = -514 kJ/mol
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The percent yield of the reaction : 89.14%
<h3>Further explanation</h3>Reaction of Ammonia and Oxygen in a lab :
<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>
mass NH₃ = 80 g
mol NH₃ (MW=17 g/mol):
mass O₂ = 120 g
mol O₂(MW=32 g/mol) :
Mol ratio of reactants(to find limiting reatants) :
mol of H₂O based on O₂ as limiting reactants :
mol H₂O :
mass H₂O :
4.5 x 18 g/mol = 81 g
The percent yield :
Charles law gives the relationship between temperature of gas and volume of gas.
It states that for a fixed amount of gas, temperature is directly proportional to volume of gas.
V / T = k
where V- volume , T - temperature and k - constant
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation.
T1 = 250 °C + 273 = 523 K
T2 = 150 °C + 273 = 423 K
Substituting the values in the equation,
V = 251 mL
the new volume is 251 mL
It states that for a fixed amount of gas, temperature is directly proportional to volume of gas.
V / T = k
where V- volume , T - temperature and k - constant
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation.
T1 = 250 °C + 273 = 523 K
T2 = 150 °C + 273 = 423 K
Substituting the values in the equation,
V = 251 mL
the new volume is 251 mL