At 298 K,what is the value of Delta H - Delt | vedclass

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(B) The chemical reaction is: $C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$The change in the number of moles of gaseous species is $\Delta n_g

Vedclass · Accepted Answer

$1238.78$

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(B) The chemical reaction is: $C_{(s)} + \frac{1}{2} O_{2(g)} ightarrow CO_{(g)}$The change in the number of moles of gaseous species is $\Delta n_g = n_{p(g)} - n_{r(g)} = 1 - 0.5 = 0.5 \ mol$.From the first law of thermodynamics,the relationship between enthalpy change and internal energy change is $\Delta H = \Delta E + \Delta n_g RT$.Therefore,$\Delta H - \Delta E = \Delta n_g RT$.Substituting the values: $\Delta H - \Delta E = 0.5 	imes 8.314 	imes 298 = 1238.78 \ J \ mol^{-1}$.

At $298 \ K$,what is the value of $\Delta H - \Delta E$ in $J \ mol^{-1}$ when $1 \ mol$ of carbon monoxide gas is formed from its elements in their standard states? $[R = 8.314 \ J \ K^{-1} \ mol^{-1}]$

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