Answer:
-892.1 kJ/mol is the change in enthalpy for this reaction.
Explanation:
The heat required to melt the ice or heat of fusion =[tex]\Delta H_{fus}=333.5 J/g[/tex]
Mass of an ice = 10.7 g
Amount of heat required to melt 10.7 grams of ice at 0 degrees Celsius: Q
[tex]Q=10.7 g\times 333.5 J/g=3,568.45 J[/tex]
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex], ΔH = ?
Heat released on combustion of 0.00400 moles of methane = Q'= -Q
Q' = -3,568.45 J = -3.568 kJ ( 1J = 0.001 kJ)
Change in enthalpy for this reaction = ΔH
[tex]\Delta H=\frac{Q'}{0.00400 mol}=\frac{-3.568 kJ}{0.00400 mol}=-892.1 kJ/mol[/tex]
-892.1 kJ/mol is the change in enthalpy for this reaction.