Given the thermochemical equation SO2(g) + ½ O2(g) ® SO3 (g) DH = -99.1 kJ
calculate the enthalpy change (DH) when 89.6 g of SO2 is converted to SO3
a. -69.3 kJ
b. -139 kJ
c. 69.3 kJ
d. 139 kJ
e. -111 kJ
Assistance needed.
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To calculate the enthalpy change (ΔH) when a certain amount of substance is converted to another, we can use the given thermochemical equation and the concept of stoichiometry.
In this case, the thermochemical equation states that 1 mole of sulfur dioxide (SO2) reacts with half a mole of oxygen gas (O2) to form 1 mole of sulfur trioxide (SO3) with a enthalpy change of -99.1 kJ.
To find the answer, we need to follow these steps:
Step 1: Calculate the number of moles of SO2 from the given mass.
To do this, we use the molar mass of SO2, which is 64.06 g/mol.
Number of moles of SO2 = mass / molar mass = 89.6 g / 64.06 g/mol
Step 2: Determine the stoichiometry ratio between SO2 and SO3.
From the balanced equation, we can see that 1 mole of SO2 reacts to produce 1 mole of SO3. Therefore, the stoichiometry ratio is 1:1.
Step 3: Calculate the enthalpy change when the given amount of SO2 is converted to SO3.
Since the stoichiometry ratio is 1:1, the enthalpy change is directly proportional to the number of moles of SO2.
ΔH = -99.1 kJ/mol x (number of moles of SO2)
Step 4: Plug in the values and calculate.
ΔH = -99.1 kJ/mol x (89.6 g / 64.06 g/mol)
ΔH = -138.95 kJ
Therefore, the enthalpy change (ΔH) when 89.6 g of SO2 is converted to SO3 is approximately -139 kJ.
So, the correct answer is (b) -139 kJ.