Question

The reaction of hydrogen sulfide(g) with oxygen(g) to form water(l) and sulfur dioxide(g) proceeds as follows: 2H2S(g) 3O2(g)2H2O(l) 2SO2(g) When 9.82 g H2S(g) reacts with sufficient O2(g), 161 kJ is evolved. Calculate the value of rH for the chemical equation given. kJ/mol

Answer 1

Answer: The value of
`\Delta H` is 1118 .

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Moles of solute}=\frac{\text{given mass}}{\text{molar mass}}`

`\text{Moles of} H_2S}=(9.82g)/(34.1g/mol)=0.288mol`

The balanced chemical reaction is:

`2H_2S(g)+3O_2(g)\rightarrow 2H_2O(l)+2SO_2(g)`

Given :

Energy released when 0.288 mole of
`H_2S` is reacted = 161 kJ

Thus Energy released when 2 moles of
`H_2S` is reacted =
`(161)/(0.288)* 2=1118kJ`

Thus the value of
`\Delta H` is 1118.