Question

What is the enthalpy change \((in~J~mol^{-1})\) for the conversion of 1 mole of \(H_{2}O (l)\) at \(10^{\circ}C\) to 1 mole of \(H_{2}O (s)\) at \(-10^{\circ}C\)? \((At~0^{\circ}C~H_{2}O(s)+x~kj~mol^{-1}\rightarrow H_{2}O(l); C_{p}(H_{2}O(l))=yJ~mol^{-1}K^{-1}; C_{p}(H_{2}O(s))=zJ~mol^{-1}K^{-1})\)

• A. \(-(1000x+10y+10z)\)
• B. \(-(x + y + z)\)
• C. \(-(1000x+y-z)\)
• D. \(-(1000x-y+z)\)

Hint:

Always ensure energy units are consistent (convert kJ to J) and pay close attention to the sign of \(\Delta T\) and the phase change enthalpy when dealing with endothermic or exothermic processes.

Answer 1

The Correct Option is A

Step 1: Use Hess's law.
The total heat change is the same no matter the path. So break the cooling and freezing into three easy steps and add them.
Step 2: Cool liquid water from 10 C to 0 C.
Heat change $=nC_p(l)\Delta T=1\times y\times(0-10)=-10y$ J. The minus sign shows heat leaves.
Step 3: Freeze water at 0 C.
Melting needs $+x$ kJ, so freezing (the reverse) releases the same amount: $\Delta H_2=-x$ kJ $=-1000x$ J.
Step 4: Cool ice from 0 C to minus 10 C.
Heat change $=nC_p(s)\Delta T=1\times z\times(-10-0)=-10z$ J.
Step 5: Add the three parts.
\[ \Delta H=-10y-1000x-10z \]
Step 6: Factor neatly.
\[ \Delta H=-(1000x+10y+10z)\ \text{J mol}^{-1} \] \[ \boxed{-(1000x+10y+10z)} \]