Question

The enthalpies of combustion of S, SO2,and H2 are-298.2, -98.7 and -287.3 kJ mol-" respectively. If
enthalpy of the reaction SO3+H2O→H2SO4 is-130.2 kJ mol, the enthalpy of formation of
H2SO4 is

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Answer 1

Answer:

-814.4 kj mol^-1

Explanation:

above pic gives you correct explanation

Answer 2

Given,

The enthalpy of combustion of $S$= $-298.2kJmol^{-1}$

The enthalpy of combustion of $SO_{2}=-98.7kJmol^{-1}$

The enthalpy of combustion of $H_{2}= -287.3kJmol^{-1}$

The enthalpy of reaction $SO_{3}+H_{2}O\rightarrow H_{2} SO_{4} = -130.2 kJ mol^{-1}$

The enthalpy of formation of $H_{2}SO_{4}$ can be given by the equation:

$S+H_{2}+2O_{2} \rightarrow H_{2} SO_{4}$

The $\Delta H _{formation}$ for $H_{2} SO_{4}$ can be calculated as follows:

$\Delta H=-(298.2+98.7+287.3+130.2)=-814.4kJmol^{-1}$

Hence, the enthalpy of formation for  $H_{2} SO_{4}$ is -814.4kJmol⁻¹.