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 .

Explanation:

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

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

$\text{Moles of} H_2S}=\frac{9.82g}{34.1g/mol}=0.288mol$    

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 = $\frac{161}{0.288}\times 2=1118kJ$

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