For the gaseous reaction A_{(g)} + 3B_{(g)} r | vedclass

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(B) The given reaction is $A_{(g)} + 3B_{(g)} \rightarrow 3C_{(g)} + 3D_{(g)}$.The change in the number of moles of gaseous products and reactants is

Vedclass · Accepted Answer

$18.2$

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(B) The given reaction is $A_{(g)} + 3B_{(g)} ightarrow 3C_{(g)} + 3D_{(g)}$.The change in the number of moles of gaseous products and reactants is $\Delta n_g = (3 + 3) - (1 + 3) = 6 - 4 = 2$.The relationship between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$.Given $\Delta U = 17 \, Kcal$,$R = 2 \, cal \, K^{-1} \, mol^{-1} = 2 	imes 10^{-3} \, Kcal \, K^{-1} \, mol^{-1}$,and $T = 27 + 273 = 300 \, K$.Substituting the values: $\Delta H = 17 + (2 	imes 2 	imes 10^{-3} 	imes 300) = 17 + 1.2 = 18.2 \, Kcal$.

For the gaseous reaction $A_{(g)} + 3B_{(g)} \rightarrow 3C_{(g)} + 3D_{(g)}$,the value of $\Delta U = 17 \, Kcal$ at $27 \, ^oC$. Assuming $R = 2 \, cal \, K^{-1} \, mol^{-1}$,the value of $\Delta H$ for the above reaction will be ....... $Kcal$.

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