If 38.55 kJ of heat is absorbed when 6.0 g | vedclass

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Question

(B) The balanced chemical equation is: $2ClF_{(g)} + O_{2(g)} \longrightarrow Cl_2O_{(g)} + OF_{2(g)}$.Given that $6.0 \ g$ of $O_2$ corresponds to $n

Vedclass · Accepted Answer

$205.6$

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(B) The balanced chemical equation is: $2ClF_{(g)} + O_{2(g)} \longrightarrow Cl_2O_{(g)} + OF_{2(g)}$.Given that $6.0 \ g$ of $O_2$ corresponds to $n = \frac{6.0 \ g}{32.0 \ g/mol} = 0.1875 \ mol$ of $O_2$.The heat absorbed for $0.1875 \ mol$ of $O_2$ is $\Delta H = +38.55 \ kJ$.The standard enthalpy of reaction $(\Delta H^0)$ is defined for the reaction as written,which involves $1 \ mol$ of $O_2$.Therefore,$\Delta H^0 = \frac{38.55 \ kJ}{0.1875 \ mol} = 205.6 \ kJ/mol$.

If $38.55 \ kJ$ of heat is absorbed when $6.0 \ g$ of $O_2$ reacts with $ClF$ according to the reaction $2ClF_{(g)} + O_{2(g)} \longrightarrow Cl_2O_{(g)} + OF_{2(g)}$. What is the standard enthalpy of reaction (in $kJ$)?

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