A solution of 500 mL of 0.2 M KOH and 500 | vedclass

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Question

(A) The reaction between a strong acid $(HCl)$ and a strong base $(KOH)$ is a neutralization reaction,which is exothermic. The heat evolved $(Q)$ is g

Vedclass · Accepted Answer

$T_1 = T_2$

Vedclass · Answer

(A) The reaction between a strong acid $(HCl)$ and a strong base $(KOH)$ is a neutralization reaction,which is exothermic. The heat evolved $(Q)$ is given by $Q = m 	imes c 	imes \Delta T$,where $m$ is the total mass of the solution,$c$ is the specific heat capacity,and $\Delta T$ is the rise in temperature.In the first case,total volume $= 500 \ mL + 500 \ mL = 1000 \ mL$. Assuming density $\approx 1 \ g/mL$,$m_1 = 1000 \ g$. Thus,$Q_1 = 1000 	imes c 	imes T_1$.In the second case,total volume $= 250 \ mL + 250 \ mL = 500 \ mL$. Thus,$m_2 = 500 \ g$. The amount of reactants is halved,so the heat evolved is also halved: $Q_2 = \frac{1}{2} Q_1 = 500 	imes c 	imes T_2$.Equating the expressions: $500 	imes c 	imes T_2 = \frac{1}{2} 	imes (1000 	imes c 	imes T_1)$.Simplifying,$500 	imes c 	imes T_2 = 500 	imes c 	imes T_1$,which gives $T_1 = T_2$.

List-$I$	List-$II$
$A$. Spontaneous process	$I$. $\Delta H < 0$
$B$. Process with $\Delta P = 0; \Delta T = 0$	$II$. $\Delta G_{T, P} < 0$
$C$. $\Delta H_{reaction}$	$III$. Isothermal and isobaric process
$D$. Exothermic process	$IV$. [Bond energies of molecules in reactants] - [Bond energies of product molecules]

$A$ solution of $500 \ mL$ of $0.2 \ M \ KOH$ and $500 \ mL$ of $0.2 \ M \ HCl$ is mixed and stirred; the rise in temperature is $T_1$. The experiment is repeated using $250 \ mL$ each of solution,the temperature raised is $T_2$. Which of the following is true?

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