For combustion of one mole of magnesium in an | vedclass

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(B) The combustion reaction of magnesium is: $Mg(s) + \frac{1}{2}O_{2}(g) \rightarrow MgO(s)$The change in the number of gaseous moles is $\Delta n_{g

Vedclass · Accepted Answer

$600$

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(B) The combustion reaction of magnesium is: $Mg(s) + \frac{1}{2}O_{2}(g) ightarrow MgO(s)$The change in the number of gaseous moles is $\Delta n_{g} = n_{p(g)} - n_{r(g)} = 0 - \frac{1}{2} = -0.5 \ mol$.The relationship between enthalpy change and internal energy change is given by: $\Delta H = \Delta U + \Delta n_{g}RT$.Given $\Delta H = -601.70 \ kJ \ mol^{-1}$,$R = 8.3 \ J \ K^{-1} \ mol^{-1} = 8.3 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$,and $T = 300 \ K$.Substituting the values: $-601.70 = \Delta U + (-0.5) 	imes (8.3 	imes 10^{-3}) 	imes 300$.$-601.70 = \Delta U - 1.245$.$\Delta U = -601.70 + 1.245 = -600.455 \ kJ \ mol^{-1}$.The magnitude of the change in internal energy is $|\Delta U| = 600.455 \ kJ \ mol^{-1}$.Rounding to the nearest integer,we get $600 \ kJ$.

For combustion of one mole of magnesium in an open container at $300 \ K$ and $1 \ bar$ pressure,$\Delta_{C}H^{\ominus} = -601.70 \ kJ \ mol^{-1}$,the magnitude of change in internal energy for the reaction is $.... \ kJ$. (Nearest integer)
(Given : $R = 8.3 \ J \ K^{-1} \ mol^{-1}$ )

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For combustion of one mole of magnesium in an open container at $300 \ K$ and $1 \ bar$ pressure,$\Delta_{C}H^{\ominus} = -601.70 \ kJ \ mol^{-1}$,the magnitude of change in internal energy for the reaction is $.... \ kJ$. (Nearest integer)(Given : $R = 8.3 \ J \ K^{-1} \ mol^{-1}$ )