For a gaseous reaction,A_{(g)} + 3B_{(g)} to | vedclass

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(B) The relationship between $\Delta H$ and $\Delta E$ is given by the equation: $\Delta H = \Delta E + \Delta n_{g} RT$. First,calculate the change i

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$18.2$

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(B) The relationship between $\Delta H$ and $\Delta E$ is given by the equation: $\Delta H = \Delta E + \Delta n_{g} RT$. First,calculate the change in the number of gaseous moles,$\Delta n_{g}$: $\Delta n_{g} = (3 + 3) - (1 + 3) = 6 - 4 = 2$. Given: $\Delta E = 17 \ kcal$,$R = 2 	imes 10^{-3} \ kcal \ K^{-1} \ mol^{-1}$,$T = 27 + 273 = 300 \ K$. Substituting the values into the equation: $\Delta H = 17 + (2 	imes 2 	imes 10^{-3} 	imes 300) = 17 + 1.2 = 18.2 \ kcal$.

For a gaseous reaction,$A_{(g)} + 3B_{(g)} \to 3C_{(g)} + 3D_{(g)}$. The value of $\Delta E$ is $17 \ kcal$ at $27 \ ^oC$. The value of $\Delta H$ in $kcal$ will be:

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