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Luden [163]
2 years ago
10

How much energy (kJ) is required to change 0.18 mole of ice (s) at 0 C to water (l) at 0 C?

Chemistry
1 answer:
Dmitriy789 [7]2 years ago
6 0

Answer:25,06 kJ of energy must be added to a 75 g block of ice.

ΔHfusion(H₂O) = 6,01 kJ/mol.

T(H₂O) = 0°C.

m(H₂O) = 75 g.

n(H₂O) = m(H₂O) ÷ M(H₂O).

n(H₂O) = 75 g ÷ 18 g/mol.

n(H₂O) = 4,17 mol.

Q = ΔHfusion(H₂O) · n(H₂O)

Q = 6,01 kJ/mol · 4,17 mol

Q = 25,06 kJ.

Explanation:

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An ideal gas sealed in a rigid 4.86-L cylinder, initially at pressure Pi=10.90 atm, is cooled until the pressure in the cylinder
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\Delta H=-11897J

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Hello,

In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:

\Delta H=\Delta U+V\Delta P

Whereas the change in the internal energy is computed by:

\Delta U=nCv\Delta T

So we compute the initial and final temperatures for one mole of the ideal gas:

T_1= \frac{P_1V}{nR}=\frac{10.90atm*4.86L}{0.082*n}=\frac{646.02K  }{n} \\\\T_2= \frac{P_2V}{nR}=\frac{1.24atm*4.86L}{0.082*n}=\frac{73.49K  }{n}

Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:

\Delta U=1mol*\frac{3}{2} (8.314\frac{J}{mol*K} )*(73.49K-646.02K )=-7140J

Then, the volume-pressure product in Joules:

V\Delta P=4.86L*\frac{1m^3}{1000L} *(1.24atm-10.90atm)*\frac{101325Pa}{1atm} \\\\V\Delta P=-4756.96J

Finally, the change in the enthalpy for the process:

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