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(A) $1$. Calculate moles of $HCl$: $n_{HCl} = \frac{0.01 	imes 25}{1000} = 2.5 	imes 10^{-4} \ mol$. Since $HCl$ is a strong monoprotic acid,$n_{H^+

Vedclass · Accepted Answer

$0.003425$

Vedclass · Answer

(A) $1$. Calculate moles of $HCl$: $n_{HCl} = \frac{0.01 	imes 25}{1000} = 2.5 	imes 10^{-4} \ mol$. Since $HCl$ is a strong monoprotic acid,$n_{H^+} = 2.5 	imes 10^{-4} \ mol$.$2$. Calculate moles of $Ca(OH)_2$: $n_{Ca(OH)_2} = \frac{0.01 	imes 50}{1000} = 5.0 	imes 10^{-4} \ mol$. Since $Ca(OH)_2$ is a strong base,$n_{OH^-} = 2 	imes 5.0 	imes 10^{-4} = 1.0 	imes 10^{-3} \ mol$.$3$. Determine the limiting reagent: $H^+$ is the limiting reagent because $2.5 	imes 10^{-4} $4$. Calculate enthalpy change: $\Delta H = n_{H^+} 	imes \Delta H^o_{neutralization} = 2.5 	imes 10^{-4} \ mol 	imes (-13.7 \ kcal \ mol^{-1}) = -3.425 	imes 10^{-3} \ kcal = -0.003425 \ kcal$.

Condition	Temperature
$(1)$. Spontaneous	$(p)$. $273 \ K$
$(2)$. At equilibrium	$(q)$. $260 \ K$
$(3)$. Non-spontaneous	$(r)$. $280 \ K$

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Match the conditions and the temperature for the 'fusion' process of $H_2O_{(s)}$ at $1 \ atm$ pressure.

Condition Temperature

$(1)$. Spontaneous $(p)$. $273 \ K$

$(2)$. At equilibrium $(q)$. $260 \ K$

$(3)$. Non-spontaneous $(r)$. $280 \ K$

From the following data,the enthalpy of dilution of $KCl$ will be ............ $kJ$.
$KCl_{(s)} + 20H_2O \to KCl(20H_2O); \Delta H = + 15.90 \ kJ$
$KCl_{(s)} + 200H_2O \to KCl(200H_2O); \Delta H = + 18.58 \ kJ$

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