Question

The amount of heat released, when $20mL$ of $0.5M$ $NaOH$ is mixed with $100mL$ of $0.1M$ $HCl$ is $x$ $kJ$. The best of neutralisation (in $kJ{mol}^{-1}$) is:

• A. $-100x$
• B. $-50x$
• C. $+100x$
• D. $+50x$

Answer 1

The correct option is A $-100x$

The number of moles of NaOH $=\frac{\text{20 mL}}{\text{1000 mL/L}}\times \text{0.5 mol/L}=\text{0.01 mol}$.

The number of moles of HCl $=\frac{\text{100 mL}}{\text{1000 mL/L}}\times \text{0.1 mol/L}=\text{0.01 mol}$.

When 0.01 moles of NaOH neutralises 0.01 moles of HCl, the heat released is x kJ.

When 1 mole of NaOH neutralizes 1 mole of HCl, heat released will be $\frac{\text{x kJ}}{\text{0.01 mol}}=\text{100x kJ/mol}$

But since heat is released, negative sign will be introduced. Hence, the answer is

$-\text{100x kJ/mol}$.