A_{(g)} to 2B_{(g)}; Delta_rH = 20 kJ/mol at | vedclass

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(B) According to Kirchhoff's law,the variation of enthalpy of reaction with temperature is given by: $\Delta_r H_2 - \Delta_r H_1 = \Delta_r C_p (T_2

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$\Delta H$ will decrease on increasing temperature

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(B) According to Kirchhoff's law,the variation of enthalpy of reaction with temperature is given by: $\Delta_r H_2 - \Delta_r H_1 = \Delta_r C_p (T_2 - T_1)$.First,calculate $\Delta_r C_p$ for the reaction $A_{(g)} 	o 2B_{(g)}$:$\Delta_r C_p = \sum 
u_p C_{p,m}(	ext{products}) - \sum 
u_r C_{p,m}(	ext{reactants})$$\Delta_r C_p = [2 	imes C_{p,m}(B)] - [1 	imes C_{p,m}(A)]$$\Delta_r C_p = [2 	imes 5 \ J \ K^{-1} \ mol^{-1}] - [20 \ J \ K^{-1} \ mol^{-1}] = 10 - 20 = -10 \ J \ K^{-1} \ mol^{-1}$.Since $\Delta_r C_p  T_1$ (increasing temperature).Therefore,$\Delta_r H_2 = \Delta_r H_1 + \Delta_r C_p (T_2 - T_1)$ will be less than $\Delta_r H_1$.Thus,$\Delta H$ will decrease on increasing temperature.

$A_{(g)} \to 2B_{(g)}$; $\Delta_rH = 20 \ kJ/mol$ at $300 \ K$. If $C_{p,m}(A) = 20 \ J \ K^{-1} \ mol^{-1}$ and $C_{p,m}(B) = 5 \ J \ K^{-1} \ mol^{-1}$,which of the following statements is correct?

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