The data shown are needed for this question.
\(\Delta H_f^\circ(\mathrm{P_4O_{10}(s)})=-3012\,\mathrm{kJ\,mol^{-1}}\)
\(\Delta H_f^\circ(\mathrm{H_2O(l)})=-286\,\mathrm{kJ\,mol^{-1}}\)
\(\Delta H_f^\circ(\mathrm{H_3PO_4(s)})=-1279\,\mathrm{kJ\,mol^{-1}}\)
What is \(\Delta H^\circ\) for the reaction shown?
\(\mathrm{P_4O_{10}(s)+6H_2O(l)\rightarrow4H_3PO_4(s)}\)
Correct Answer: \( \boxed{\mathrm{B}} \)
Use standard enthalpies of formation:
\(\Delta H^\circ=\sum\Delta H_f^\circ(\text{products})-\sum\Delta H_f^\circ(\text{reactants})\)
Products:
\(\mathrm{4(-1279)=-5116\,kJ\,mol^{-1}}\)
Reactants:
\(\mathrm{-3012+6(-286)=-4728\,kJ\,mol^{-1}}\)
Therefore,
\(\mathrm{\Delta H^\circ=-5116-(-4728)=-388\,kJ\,mol^{-1}}\)
Therefore, the correct answer is (B).
Some standard enthalpy of combustion data are given.
Using these data, what is the enthalpy change of formation of methanol?
Correct Answer: \( \boxed{\mathrm{A}} \)
Apply Hess’ Law.
Formation reaction of methanol:
\( \mathrm{C(s) + 2H_2(g) + \dfrac{1}{2}O_2(g) \rightarrow CH_3OH(l)} \)
Using combustion data:
\( \Delta H_f^\circ = \Delta H_c^\circ(\mathrm{C}) + 2\Delta H_c^\circ(\mathrm{H_2}) – \Delta H_c^\circ(\mathrm{CH_3OH}) \)
Substitute the values:
\( \Delta H_f^\circ = (-394) + 2(-286) – (-726) \)
\( \Delta H_f^\circ = -394 -572 +726 \)
\( \Delta H_f^\circ = -240\,\mathrm{kJ\,mol^{-1}} \)
Therefore, the correct answer is (A).
Carbon monoxide and methanol can react together to form ethanoic acid.
\( \mathrm{CO(g)+CH_3OH(l)\rightarrow CH_3CO_2H(l)} \qquad \Delta H^\circ_{\mathrm{r}} \)
Standard enthalpy changes of combustion are given in the table.
| Compound | Standard enthalpy change of combustion, \( \Delta H^\circ_{\mathrm{c}} \) |
|---|---|
| CO | \(-283.0\,\mathrm{kJ\,mol^{-1}}\) |
| \( \mathrm{CH_3OH} \) | \(-726.0\,\mathrm{kJ\,mol^{-1}}\) |
| \( \mathrm{CH_3CO_2H} \) | \(-874.1\,\mathrm{kJ\,mol^{-1}}\) |
What is the value of \( \Delta H^\circ_{\mathrm{r}} \) for the reaction between carbon monoxide and methanol?
Correct Answer: \( \boxed{\mathrm{B}} \)
Using Hess’s Law,
\( \Delta H^\circ_{\mathrm{r}}=\sum \Delta H^\circ_{\mathrm{c}}(\text{reactants})-\sum \Delta H^\circ_{\mathrm{c}}(\text{products}) \)
Reactants:
\( -283.0+(-726.0)=-1009.0\,\mathrm{kJ\,mol^{-1}} \)
Products:
\( -874.1\,\mathrm{kJ\,mol^{-1}} \)
Therefore,
\( \Delta H^\circ_{\mathrm{r}}=-1009.0-(-874.1)=-134.9\,\mathrm{kJ\,mol^{-1}} \)
Hence, the correct answer is (B).
