A solution of 500 mL of 2 M KOH is added t | vedclass

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Question

(A) The reaction is $KOH + HCl \longrightarrow KCl + H_2O$. The heat evolved $(Q)$ is proportional to the number of moles of $H^+$ and $OH^-$ reacting

Vedclass · Accepted Answer

$T_1 = T_2$

Vedclass · Answer

(A) The reaction is $KOH + HCl \longrightarrow KCl + H_2O$. The heat evolved $(Q)$ is proportional to the number of moles of $H^+$ and $OH^-$ reacting. In the first case,moles of $KOH = 0.5 \ L 	imes 2 \ M = 1 \ mol$ and moles of $HCl = 1 \ mol$. Total heat $Q_1 = 1 \ mol 	imes \Delta H_{neutralization}$. In the second case,moles of $KOH = 0.25 \ L 	imes 2 \ M = 0.5 \ mol$ and moles of $HCl = 0.5 \ mol$. Total heat $Q_2 = 0.5 \ mol 	imes \Delta H_{neutralization}$. Thus,$Q_1 = 2Q_2$. The heat absorbed by the solution is given by $Q = m 	imes c 	imes \Delta T$,where $m$ is the mass of the solution. For the first case,$Q_1 = (1000 \ g) 	imes c 	imes T_1$. For the second case,$Q_2 = (500 \ g) 	imes c 	imes T_2$. Since $Q_1 = 2Q_2$,we have $1000 	imes c 	imes T_1 = 2 	imes (500 	imes c 	imes T_2)$. This simplifies to $1000 	imes T_1 = 1000 	imes T_2$,which means $T_1 = T_2$.

Process	Entropy Change
$(a)$ Liquid to vapor conversion	$(1)$ $\Delta S = 0$
$(b)$ Process not spontaneous at any temperature	$(2)$ $\Delta S = (+)$
$(c)$ Reversible expansion of an ideal gas	$(3)$ $\Delta S = (-)$

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