When 5 mol of reactant A reacts with 8 mol | vedclass

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(B) The balanced reaction is $3A_{(g)} + 4B_{(g)} 	o 2C_{(g)}$.Initial moles: $A = 5 \ mol$,$B = 8 \ mol$.For $A$: $5/3 \approx 1.67$. For $B$: $8/4

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$-72$

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(B) The balanced reaction is $3A_{(g)} + 4B_{(g)} 	o 2C_{(g)}$.Initial moles: $A = 5 \ mol$,$B = 8 \ mol$.For $A$: $5/3 \approx 1.67$. For $B$: $8/4 = 2$. Thus,$A$ is the limiting reagent.Since $5 \ mol$ of $A$ produces $100 \ kJ$ of heat,for $3 \ mol$ of $A$ (stoichiometric coefficient),the heat liberated $(\Delta U)$ is: $\Delta U = -(100 \ kJ / 5 \ mol) 	imes 3 \ mol = -60 \ kJ/mol$.In a rigid container,the heat liberated is equal to the change in internal energy,$\Delta U = -60 \ kJ/mol$.Using the relation $\Delta H = \Delta U + \Delta n_g RT$:$\Delta n_g = 2 - (3 + 4) = -5$.$\Delta H = -60 \ kJ/mol + (-5 \ mol) 	imes (8 \ J/mol \cdot K) 	imes (300 \ K) 	imes (10^{-3} \ kJ/J)$.$\Delta H = -60 + (-5 	imes 8 	imes 300 / 1000) = -60 - 12 = -72 \ kJ/mol$.

When $5 \ mol$ of reactant $A$ reacts with $8 \ mol$ of reactant $B$ in a closed rigid container according to the reaction:
$3A_{(g)} + 4B_{(g)} \to 2C_{(g)}$
$10^{2} \ kJ$ of heat was liberated. The $\Delta_rH$ of the given balanced reaction in $kJ/mol$ at $300 \ K$ is: $[R = 8 \ J/mol \cdot K]$

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When $5 \ mol$ of reactant $A$ reacts with $8 \ mol$ of reactant $B$ in a closed rigid container according to the reaction:$3A_{(g)} + 4B_{(g)} \to 2C_{(g)}$$10^{2} \ kJ$ of heat was liberated. The $\Delta_rH$ of the given balanced reaction in $kJ/mol$ at $300 \ K$ is: $[R = 8 \ J/mol \cdot K]$