1.A system releases 691 kJ of heat and does 100 kJ of work on the surroundings.

Question

1.A system releases 691 kJ of heat and does 100 kJ of work on the surroundings.

What is the change in internal energy of the system? Express your answer to 2 sig figs.

2.If 1.7 g of matter were converted to energy, how much energy would be formed?

Express your answer using two significant figures.

3.Hydrogen gas reacts with oxygen to form water.

2H2(g)+O2(g)2H2O(g)H=483.5kJ

Determine the minimum mass of hydrogen gas required to produce 524 kJ of heat.

Express your answer to three significant figures and include the appropriate units.

4. Calculate Hrxn for the following reaction:
5C(s)+6H2(g)C5H12(l)
Use the following reactions and given Hs.
C5H12(l)+8O2(g)5CO2(g)+6H2O(g), H= -3244.8 kJ
C(s)+O2(g)CO2(g), H= -393.5 kJ
2H2(g)+O2(g)2H2O(g), H= -483.5 kJ

Express answer to 4 sig figs.

5.

Use the H°f and H°rxn information provided to calculate H°f for IF:

Explanation / Answer

1. For the given system,

change in internal energy dE,

dE = q + w

with,

q = -691 kJ

w = -100 kJ

we get,

dE = -691 - 100 = -791 kJ

3. For the given water formation reaction

2 moles of H2 gas gives off 483.5 kJ of heat

So,

moles of hydrogen gas to give 524 kJ heat = 524 x 2/483.5 = 2.17 mols

mass of hydrogen needed to produce this much heat = 2.17 mol x 2 g/mol = 4.34 g

4. Using Hesse's law,

multiply equation 2 with 5 and 3 with 3 we get,

5C(g) + 5O2(g) --> 5CO2(g)   dH = 5 x -393.5 = -1987.5 kJ

6H2(g) + 3O2(g) --> 6H2O(g) dH = 3 x -483.5 = -1450.5 kJ

Add the two equation,

5C(g) + 6H2(g) + 8O2(g) ---> 5CO2(g) + 6H2O(g)   dH = -3438 kJ

Invert the first equation,

5CO2(g) + 6H2O(g) --> C5H12(l) + 8O2(g)   dH = 3244.8 kJ

Add the above two equations,

5C(g) + 6H2(g) ---> C5H12(l)   dH = -3438 + 3244.8 = -193.2 kJ

So, the answer dHrxn = -193.2 kJ

5. For the given reaction.,

dHrxn = (dHof[IF5] + 2 x dHof[IF]) - (dHof[IF7])

Feeding the values given above,

-89 = (-840 + 2 x dHof[IF]) - (-941)

dHof[IF] = (-89 - 941 + 840)/2 = -95 kJ/mol

Answer : d. -95 kJ/mol          

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