Hydrogen sulfide, H2S is a foul smelling gas. It burns to form sulfur dioxide.
2H2S +3O2 ----> 2N2 +6H2O Delta H=-1036kJ
Calculate the enthalpy change to burn 26.7 g of hydrogen sulfide.
Delta H for the reaction is 1036 kJ for 2*molar mass H2S. So what is delta H for 26.7 g instead of 2*32.066 g H2S?
17.4
To calculate the enthalpy change required to burn 26.7g of hydrogen sulfide (H2S), we can use the stoichiometry of the balanced equation and the given enthalpy change (∆H) value.
First, we need to determine the number of moles of H2S present in 26.7g of the compound. To do this, we use the molar mass of H2S, which can be found in the periodic table. The molar mass of hydrogen (H) is approximately 1g/mol, and sulfur (S) is approximately 32g/mol.
Molar mass of H2S = (2 x molar mass of H) + molar mass of S
= (2 x 1g/mol) + 32g/mol
= 2g/mol + 32g/mol
= 34g/mol
Next, we use the formula:
Number of moles = Mass / Molar mass
Number of moles of H2S = 26.7g / 34g/mol
= 0.785 moles
Now that we know the number of moles of H2S, we can use the stoichiometry of the balanced equation to determine the enthalpy change for this amount of substance.
From the balanced equation:
2H2S + 3O2 → 2N2 + 6H2O (∆H = -1036 kJ)
The stoichiometry indicates that two moles of H2S are required to produce 1036 kJ of heat. Therefore, we can set up a proportion:
(0.785 moles H2S / 2 moles H2S) = (∆H / -1036 kJ)
Let's solve for ∆H:
∆H = (0.785/2) x (-1036)
= -405.1 kJ
Therefore, the enthalpy change (∆H) to burn 26.7g of hydrogen sulfide is approximately -405.1 kJ. The negative sign indicates the release of heat (exothermic reaction) during the burning of hydrogen sulfide.