Question
How many molecules of $$ATP,$$ undergo hydrolysis to raise the temperature of $$180\,kg$$ of water which was originally at room temperature by $${1^ \circ }C?C\left\{ {P,m} \right\}$$ water $$ = 75.32\,J/mol/K,\,\Delta H\left\{ P \right\}$$ 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 :
$$1.5 \times {10^{25}}$$
Solution :
$$\eqalign{
& {q_p} = \Delta H = {C_p}\,dT \cr
& \Rightarrow {q_p} = 75.32\frac{J}{{K\,mol}} \times \left( {299 - 298} \right)K \cr
& \Rightarrow {q_p} = 75.32\frac{J}{{K\,mol}} \cr
& {\text{For }}180{\text{ }}kg{\text{ of water, no}}{\text{. of moles of water}} \cr
& {\text{ = }}\frac{{180 \times {{10}^3}g}}{{18g/mol}} \cr
& = {10^4}g\,moles \cr
& {q_p} = 75.32\frac{J}{{mol}} \times {10^4}\,moles \cr
& = 753.2 \times {10^3}\,J \cr
& = 753.2\,kJ \cr
& \Delta H\,{\text{for}}\,ATP = 7\,kcal/mol \cr
& = 7 \times 4.184\,kJ/mol \cr
& = 29.2\,kJ/mol \cr
& 6.022 \times {10^{23}}\,{\text{molecules of }}ATP{\text{ produce}} \cr
& = 29.2\,kJ \cr
& 29.2\,kJ\,{\text{produced from}} \cr
& 6.022 \times {10^{23}}{\text{molecules}} \cr
& 753.2\,kJ\,{\text{produced from}} \cr
& 6.022 \times {10^{23}} \times \frac{{75.8}}{{29.2}} \cr
& = 1.5 \times {10^{25}}\,{\text{molecules}} \cr} $$