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For glucose, C6H12O6(s), Hf = -1263 kJ/mol. Calculate the
Chapter 16, Problem 16.1.29(choose chapter or problem)
For glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s), \Delta H_{f}=-1263 \mathrm{~kJ} / \mathrm{mol}\). Calculate the enthalpy change when 1 mol of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)\) combusts to form \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\).
Questions & Answers
QUESTION:
For glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s), \Delta H_{f}=-1263 \mathrm{~kJ} / \mathrm{mol}\). Calculate the enthalpy change when 1 mol of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)\) combusts to form \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\).
ANSWER:Step 1 of 3
Chemical reaction for when 1 mol of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)\) combusts to form \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) is
\(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \rightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)\)
we are asked to calculate the enthalpy change of combustion reaction of 1 mol glucose and given
\(\Delta \mathrm{H}_{f}(\text { glucose })=-1263.00 \mathrm{~kJ} / \mathrm{mol}\)