Hydrogen sulfide, H2S, is a foul-smelling gas. It burns to form sulfur dioxide.
2H2S9(g) + 3O2(g) → 2SO2(g) + 2H2O(g); ∆H = -1037kJ
Calculate the enthalpy change to burn 36.9g of hydrogen sulfide
I corrected a typo where you typed a 9 before the (.
2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g) ∆H = -1037kJ
Calculate the enthalpy change to burn 36.9g of hydrogen sulfide
So dH is -1037 kJ for 2*(2 + 32) = 2*34 = 68 g H2S. Easy to convert to 36.9 g this way. -1037 kJ x ((36.9/68) = ? OR you can set up a proportion like this.
(-1037 kJ/68 g) = (x kJ/36.9 g) and solve for x
Well, well, well... looks like we have a chemistry question here! Let's get our burn on.
First things first, we need to convert the amount of hydrogen sulfide from grams to moles. The molar mass of hydrogen sulfide (H2S) is 34.08 g/mol (2*1.01 g/mol for hydrogen + 32.06 g/mol for sulfur).
36.9 g H2S * (1 mol/34.08 g) = 1.08 mol H2S
Now that we have the number of moles of H2S, we can use the balanced equation to determine the stoichiometric ratio between H2S and SO2, and calculate the enthalpy change.
From the balanced equation, we can see that 2 moles of H2S reacts to produce 2 moles of SO2, releasing 1037 kJ of heat.
So, for 1.08 mol of H2S, the enthalpy change will be:
1.08 mol H2S * (-1037 kJ/2 mol H2S) = -557.4 kJ
Voilà! The enthalpy change to burn 36.9g of hydrogen sulfide is approximately -557.4 kJ.
Keep in mind that this calculated enthalpy change assumes that the reaction is taking place under standard conditions. You know, the kind of conditions you can find in a chemistry textbook or a lab manual. Just making sure we're all on the same page!
To calculate the enthalpy change to burn 36.9g of hydrogen sulfide, we need to use the given molar mass and stoichiometry of the reaction.
The molar mass of hydrogen sulfide (H2S) can be calculated as follows:
2(1.008 g/mol) + 32.07 g/mol = 34.076 g/mol
Now we can calculate the number of moles of hydrogen sulfide in 36.9g:
36.9 g ÷ 34.076 g/mol ≈ 1.0832 mol
According to the balanced equation, the stoichiometric ratio between H2S and ∆H is 2:1037 kJ. Therefore, we can calculate the enthalpy change using dimensional analysis:
∆H = (1.0832 mol H2S) × (-1037 kJ / 2 mol H2S)
∆H ≈ -559.869 kJ
Hence, the enthalpy change to burn 36.9g of hydrogen sulfide is approximately -559.869 kJ.
To calculate the enthalpy change to burn 36.9g of hydrogen sulfide (H2S), we first need to determine the number of moles of H2S. We can use the molar mass of H2S to convert grams to moles.
The molar mass of H2S is:
2(atomic mass of hydrogen) + 1(atomic mass of sulfur)
= 2(1.01 g/mol) + 32.07 g/mol
= 34.09 g/mol
Now we can calculate the number of moles:
moles of H2S = mass of H2S / molar mass of H2S
= 36.9 g / 34.09 g/mol
≈ 1.081 mol
Next, we can use the balanced chemical equation to determine the molar ratio between H2S and SO2. From the equation:
2H2S + 3O2 → 2SO2 + 2H2O
We can see that the ratio is 2:2, meaning that 2 moles of H2S will produce 2 moles of SO2.
Therefore, the number of moles of SO2 produced will also be 1.081 mol.
We can now calculate the enthalpy change by multiplying the number of moles of H2S by the enthalpy change per mole, which is -1037 kJ/mol (from the given value in the question).
Enthalpy change = moles of H2S x enthalpy change per mole
= 1.081 mol x -1037 kJ/mol
≈ -1120.997 kJ
So, the enthalpy change to burn 36.9g of hydrogen sulfide is approximately -1120.997 kJ.