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  • Calculate the standard enthalpy of combustion (kJ/mol) of glucose Given, <img alt="\Delta H^0_fCO_2 = -394\ kJ/mol" src="http://entrancecorner.oncodecogs.com/gif.latex?%5CDelta%20H%5

Calculate the standard enthalpy of combustion (kJ/mol) of glucose

Given, \Delta H^0_fCO_2 = -394\ kJ/mol, \Delta H^0_fH_2O = -286\ kJ/mol, \Delta H^0_fC_6H_{12}O_6 = -1275\ kJ/mol

 

Option: 1

2805


Option: 2

-2805


Option: 3

1275


Option: 4

-1275


Answers (1)

best_answer

 

\Delta H^o_f {O_2} is equal to Zero because O2 is in the standard state.

First, write the balanced equation of the reaction,

C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O

\Delta H_{c} = \left (6*\Delta H_{f}^{0}H_{2}0 + 6*\Delta H_{f}^{0}CO_{2} \right )-\left ( \Delta H_{f}^{0}C_{6}H_{12}O_{6} + 6*\Delta H_{f}^{0}O_{2} \right )

\Delta H_c = 6\times (-394) + 6\times (-286) - (-1275) -6\times (0)

\Delta H_c = -2805\ kJ/mol

Hence, Option number (2) is correct.

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Rishi

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