## Hayatcakes 3 years ago 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

1. JFraser

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

2. Hayatcakes

there is 1.22 moles

3. JFraser

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

4. Hayatcakes

okay so the answer would −7.87 kJ ?

5. Hayatcakes

or am i wrong?

6. JFraser

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

7. Hayatcakes

8. JFraser

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 $0.767mol H_2O * (\frac{40.65kJ}{1mol H_2O}) = 31.17kJ$

9. mos1635

40.65 or 6.03 ????

10. Hayatcakes

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

11. JFraser

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

12. mos1635

-4.623$\approx$ -4.62