Question

$50.0mL$ of $0.10MHCl$ is mixed with $50.0mL$ of $0.10MNaOH$. The solution's temperature rises by ${3.0}^{\circ }C$. Calculate the enthalpy of neutralization per mole of $HCl$. (Assuming density of sol. $=1g/ml$ and specifice heat of water).

• A. $-2.5\times {10}^{2}kJ/mole$
• B. $-1.3\times {10}^{2}kJ/mole$
• C. $-8.4\times {10}^{1}kJ/mole$
• D. $-6.3\times {10}^{1}kJ/mole$

Answer 1

The correct option is A $-2.5\times {10}^{2}kJ/mole$

$HCl+NaOH\to NaCl+H_{2}O$

Molarity$=\frac{no.ofmoles\left(n\right)}{volume\left(L\right)}$

No. of moles of acid used$=0.05\times 0.1=0.05$

No. of moles of base used$=0.05\times 0.1=0.05$

So, $0.05$ moles of $HCl$ reacts with $0.05$ moles of $NaOH$ to form $0.05$ moles of $NaCl$.

$△H=-mC△T$

$m=massofsolution=volume\times density$

Total volume$=50ml+50ml=100ml$

$m=100\times 1=100g=0.1kg$

$△H=-(0.1\times 4.18\times 3)$ $[C_{water}=4.18\frac{kJ}{K{g}^{\circ }C}]$

$△H=-1.254kJ$

$0.005moles\to -1.254kJ$

$1mole\to \frac{-1.254}{0.05}=-250.8kJ/mole$