If Delta E is the heat of reaction for C_2H_5 | vedclass

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(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by the equation: $\Delta H = \Delta E + \De

Vedclass · Accepted Answer

$\Delta H = \Delta E - RT$

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(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by the equation: $\Delta H = \Delta E + \Delta n_g RT$.Here,$\Delta n_g$ is the change in the number of moles of gaseous species,calculated as: $\Delta n_g = \sum n_{g, 	ext{products}} - \sum n_{g, 	ext{reactants}}$.For the reaction $C_2H_5OH_{(l)} + 3O_{2(g)} \longrightarrow 2CO_{2(g)} + 3H_2O_{(l)}$,the gaseous moles are:Products: $2 	ext{ moles of } CO_2$.Reactants: $3 	ext{ moles of } O_2$.$\Delta n_g = 2 - 3 = -1$.Substituting this value into the equation,we get: $\Delta H = \Delta E + (-1)RT = \Delta E - RT$.

If $\Delta E$ is the heat of reaction for $C_2H_5OH_{(l)} + 3O_{2(g)} \longrightarrow 2CO_{2(g)} + 3H_2O_{(l)}$ at constant volume,the $\Delta H$ (heat of reaction at constant pressure) at constant temperature is

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