1. How much heat is needed to warm 250.4 G of water from 22.6 degrees C to 75.8
ID: 856353 • Letter: 1
Question
1. How much heat is needed to warm 250.4 G of water from 22.6 degrees C to 75.8 degrees C given that the specific heat capacity of water is 4.184 Joules divided by Grams multiplied by degree C?
2. Given that the following reaction and enthalpy change, calculate the amount of heat required to convert 100.4g of liquid water (18.02 g/mole) to vapor? H2O(l)-->H2O(g) deltaH = +44 kJ
3. Given the following reactions N2(g) + O2(g) ----> 2NO(g) delta H = +180.7 kJ / 2NO(g) + O2(g) ----> 2NO2(g) delta H = -113.1 kJ
What is the enthaply of the reaction for 4NO(g) ----> 2NO2(g) + N2(g)
Is the reaction endothemic or exothermic?
4. When 0.800 grams of NaOH is dissolved in 100.0 grams of water, the temperature of the solution increases from 25.00 degrees C to 27.06 degrees C. Calculate delta H in kJ/mol using the correct sign. The molar mass of NaOH is 40.00 g/mol and the specific heat of water is 4.18 J/g minus degrees C.
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Explanation / Answer
2. Given that the following reaction and enthalpy change, calculate the amount of heat required to convert 100.4g of liquid water (18.02 g/mole) to vapor? H2O(l)-->H2O(g) deltaH = +44 kJ
H2O(l)-->H2O(g) deltaH = +44 kJ
18.02g H2O(l)-->18.02gH2O(g) deltaH = +44 kJ multiply all terns by 5.572
100.4g H2O(l)-->100.4gH2O(g) deltaH = +244.17 kJ
4. When 0.800 grams of NaOH is dissolved in 100.0 grams of water, the temperature of the solution increases from 25.00 degrees C to 27.06 degrees C. Calculate delta H in kJ/mol using the correct sign. The molar mass of NaOH is 40.00 g/mol and the specific heat of water is 4.18 J/g minus degrees C.
moles of NaOH=0.8g/40g/mole=0.02moles
Heat prduced by reaction=heat absorbed by water=Q=mCDeltaT=mC(T1-T2)
Q=100.8g*(4.18J/g.oC-1)*(25-27.06)oC=-867.97J=-0.868KJ
now heat produced=-0.868KJ/0.02moles=-433.98KJ/mole
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