What is the enthalpy change when 13.8 grams of H2O changes from a liquid to a solid at zero degrees Celsius? ΔHcondensation = −40.65 kJ/mol ΔHfreezing = −6.03 kJ/mol

−4.62 kJ

−83.2 kJ

−7.87 kJ

−31.3 kJ

Question

What is the enthalpy change when 13.8 grams of H2O changes from a liquid to a solid at zero degrees Celsius? ΔHcondensation = −40.65 kJ/mol     ΔHfreezing = −6.03 kJ/mol

		
−4.62 kJ
 		
−83.2 kJ
 		
−7.87 kJ
 		
−31.3 kJ

jfraser · Answer

how many moels of water is in 13.8g of H2O?

anonymous · Answer

there is 1.22 moles

jfraser · Answer

no. 1mol of H2O is 18g. You've only got 13.8g. Flip the fraction upside down

anonymous · Answer

okay so the answer would −7.87 kJ
 		?

anonymous · Answer

or am i wrong?

jfraser · Answer

your mole conversion should look like$$13.8g H_2O * (\frac{1mol H_2O}{18gH_2O}) = 0.767 moles H_2O$$

anonymous · Answer

so is my answer wrong?

jfraser · Answer

mole of water "costs" 40.65kJ of energy to freeze. You've got only 76.7% of 1 mole. The heat of fusion for .767mol of H2O will be 
$$0.767mol H_2O * (\frac{40.65kJ}{1mol H_2O}) = 31.17kJ$$

mos1635 · Answer

40.65 or 6.03 ????

anonymous · Answer

Oh okay .
−31.3 kJ so this would be the final result?

jfraser · Answer

@mos1635 is correct, replace the 40.65kJ with 6.03kJ. I missed the condensation vs freezing.

mos1635 · Answer

-4.623$$\approx$$ -4.62