Question

$50.0mL$ of $0.10MHCl$ is mixed with $50.0mL$ of $0.10MNaOH$. The solution temperature rises by $3.0C$. The enthalpy of neutralization per mol of $HCl$ is

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

Answer 1

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

$HCl+NaOH⟶NaCl+H_{2}O$
enthalpy change $=m{C}_{v}dT=100\times 4.2\times 3$
$=1.26kJ$
enthalpy change for 5 millimole $=1.26kJ$
enthalpy change for 1 mole $\Rightarrow \frac{1.26}{5\times {10}^{-3}}$
$\Rightarrow 2.52\times {10}^{2}kJ$