What is the difference between Delta H and De | vedclass

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

Vedclass · Accepted Answer

$-7.43$

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(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H - \Delta U = \Delta n_g RT$Here,$\Delta n_g$ is the change in the number of moles of gaseous species.For the reaction: $2 \ C_6H_{6(\ell)} + 15 \ O_{2(g)} ightarrow 12 \ CO_{2(g)} + 6 \ H_2O_{(\ell)}$$\Delta n_g = (n_{g, 	ext{products}}) - (n_{g, 	ext{reactants}}) = 12 - 15 = -3$Now,substituting the values: $\Delta H - \Delta U = -3 	imes 8.314 \ J \ K^{-1} \ mol^{-1} 	imes 298 \ K$$= -7432.7 \ J = -7.432 \ kJ$

What is the difference between $\Delta H$ and $\Delta U$ for the reaction given below at $298 \ K$ (in $kJ$)? $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$
$2 \ C_6H_{6(\ell)} + 15 \ O_{2(g)} \rightarrow 12 \ CO_{2(g)} + 6 \ H_2O_{(\ell)}$

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Reason : Enthalpy change $(\Delta H)$ is zero for an isothermal process.

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What is the difference between $\Delta H$ and $\Delta U$ for the reaction given below at $298 \ K$ (in $kJ$)? $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$$2 \ C_6H_{6(\ell)} + 15 \ O_{2(g)} \rightarrow 12 \ CO_{2(g)} + 6 \ H_2O_{(\ell)}$