For the reaction H_{2}F_{2(g)} rightarrow H_{ | vedclass

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(A) The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta U + \Delta n_{g} RT$.For the reaction $H_{2}F_{

Vedclass · Accepted Answer

$57$

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(A) The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta U + \Delta n_{g} RT$.For the reaction $H_{2}F_{2(g)} ightarrow H_{2(g)} + F_{2(g)}$,the change in the number of gaseous moles is $\Delta n_{g} = (1 + 1) - 1 = 1$.The temperature is $T = 27 + 273 = 300 \ K$.Given $\Delta U = -59.6 \ kJ \ mol^{-1}$ and $R = 8.314 \ J \ K^{-1} \ mol^{-1} = 8.314 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$.Substituting the values: $\Delta H = -59.6 + (1 	imes 8.314 	imes 10^{-3} 	imes 300) = -59.6 + 2.4942 = -57.1058 \ kJ \ mol^{-1}$.The nearest integer value is $-57 \ kJ \ mol^{-1}$.

For the reaction $H_{2}F_{2(g)} \rightarrow H_{2(g)} + F_{2(g)}$,$\Delta U = -59.6 \ kJ \ mol^{-1}$ at $27^{\circ} C$. The enthalpy change for the above reaction is $(-)$ $kJ \ mol^{-1}$ [nearest integer]. Given: $R = 8.314 \ J \ K^{-1} \ mol^{-1}$.

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