In an experiment on the enthalpy of neutralis | vedclass

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(B) The reaction is: $2NaOH + H_2SO_4 ightarrow Na_2SO_4 + 2H_2O$. Number of millimoles of $NaOH = 50 \ cm^3 	imes 0.4 \ M = 20 \ mmol$. Since $2 \

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(B) The reaction is: $2NaOH + H_2SO_4 ightarrow Na_2SO_4 + 2H_2O$. Number of millimoles of $NaOH = 50 \ cm^3 	imes 0.4 \ M = 20 \ mmol$. Since $2 \ mol$ of $NaOH$ react with $1 \ mol$ of $H_2SO_4$,the $20 \ mmol$ of $NaOH$ will require $10 \ mmol$ of $H_2SO_4$. Volume of $H_2SO_4$ required $= \frac{10 \ mmol}{0.25 \ M} = 40 \ cm^3$. As $H_2SO_4$ is added,the temperature of the solution increases due to the exothermic nature of the neutralisation reaction until the equivalence point is reached at $40 \ cm^3$. After the equivalence point,the addition of excess $H_2SO_4$ (which is at the same initial temperature) causes the total volume of the solution to increase,leading to a decrease in temperature due to dilution. Therefore,the temperature plot should show a maximum at $40 \ cm^3$. Plot $B$ shows the maximum temperature at $40 \ cm^3$.

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