Calculate entropy of the reaction 1/2N$_2$(g) + 3/2H$_2$(g)?NH$_3$(g) at 298 and 1000K. SÂș(NH$_3$) = 192.66 J/mol·K; SÂș(H$_2$) = 130.52 J/mol·K; c$_p$(NH$_3$) = 29.80+25.48e-3T-1.67e5T$^{-2}$ c$_p$(N$_2$) = 27.88+4.27e-3T c$_p$(H$_2$) = 27.28+33.26e-3T+0.5e5T$^{-2}$ SÂș(N$_2$) = 191.50 J/mol·K
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Step 1: The entropy change for a reaction is given by: $$\Delta S = \sum S^\circ(products) - \sum S^\circ(reactants)$$ Show moreâŠ
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