Question

How many molecules of ATP, undergo hydrolysis to raise the temperature of $180\, kg$ of water which was originally at room temperature by $1^{\circ} C$ ? $C\{P, m\}$ water $=75.32\, J / mol / K$, $\Delta H\{P\}$ for ATP hydrolysis $=7\, kcal / mol$

• A. $1.5 \times 10^{25}$
• B. $2.00 \times 10^{23}$
• C. $3.4 \times 10^{25}$
• D. $4.0 \times 10^{24}$

Answer 1

Answer: (A)

$q_p=\Delta H=C_p d T$
$\Rightarrow q_p=75.32 \frac{ J }{ \text{K mol }} \times(299-298) K $
$\Rightarrow q_p=75.32 \frac{ J }{ \text{K mol }}$
For $180 \,kg$ of water, no. of moles of water
$=\frac{180 \times 10^3 g }{18 g / mol }=10^4 \text { moles }$
$q_p=75.32 \frac{ J }{ \text{mol }} \times 10^4 \text { moles }$
$\quad=753.2 \times 10^3 J =753.2 \,kJ $
$\Delta H \text { for ATP } =7 \,kcal / mol$
$=7 \times 4.184\, kJ / mol =29.2\, kJ / mol$
$29.2 \,kJ$ produced from $6.022 \times 10^{23}$ molecules
$753.2\, kJ$ produced from $6.022 \times 10^{23} \times \frac{753.2}{29.2}$
$=1.5 \times 10^{25}$ molecules