For the complete combustion of methanolCH_3OH | vedclass

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(D) The heat measured by a bomb calorimeter is the change in internal energy,$\Delta U = -726 \ kJ \ mol^{-1}$.The reaction is $CH_3OH_{(l)} + \frac{3

Vedclass · Accepted Answer

$727$

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(D) The heat measured by a bomb calorimeter is the change in internal energy,$\Delta U = -726 \ kJ \ mol^{-1}$.The reaction is $CH_3OH_{(l)} + \frac{3}{2} O_{2(g)} ightarrow CO_{2(g)} + 2H_2O_{(l)}$.The change in the number of gaseous moles is $\Delta n_g = n_{p(g)} - n_{r(g)} = 1 - \frac{3}{2} = -0.5 \ mol$.The relationship between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$.Given $T = 27 + 273 = 300 \ K$ and $R = 8.3 \ J \ K^{-1} \ mol^{-1} = 8.3 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$.$\Delta H = -726 + (-0.5) 	imes (8.3 	imes 10^{-3}) 	imes 300$.$\Delta H = -726 - 1.245 = -727.245 \ kJ \ mol^{-1}$.Rounding to the nearest integer,$x = 727$.

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