Question

1/ the total volume of hydrogen gass needed to fill the Hindenburge was 2x10^8 liters at 1 atm and 25*C. How Much energy was evolved when it burned ?

H2(g) + .5O2(g) ----> H2O(I) ∆H* = -286Kj

2/ The heat of formation of iron(III) oxide is -826KJ/mol. Calculate the heat of the reaction when a 55.8 g sample of iron is reacted with an excess of oxygen gas.

4Fe(s) + 3O2(g) -----> 2Fe2O3(s)

3/ the vapor pressure of water at 25*Celcius is 23.8 torr. Determine the mass of glucose( molar mass= 180g/mol) needed to add to 500 grams of water to change the vapor pressure to 23.1 torr.

4/ Consider the reaction H2(g) + .5O2(g) ----> H2O(I) ∆H*= -286 Kj

which of the following is true?

A/ The reaction is exothermic

B/ The reaction is endothermic

C/ The enthalpy of the products is less than that of the reactants.

D/ Heat is absorbed by the system

E/ The reaction is exothermic and the enthalpy of the products is less than that of the reactants.

Answer 1

1/ the total volume of hydrogen gass needed to fill the Hindenburge was 2x10^8 liters at 1 atm and 25*C. How Much energy was evolved when it burned ?

H2(g) + .5O2(g) ----> H2O(I) ∆H* = -286Kj

For 1 mole we have -286 kJ

d = 0.089 g/L

moles of H2 = 2x10^8 liters x 0.89 x 1000/2.02 = 8.9 x 10^12 mols

So for 8.9 x 10^12 mols = 8.9 x 10^12 x -286 = -2.55 x 10^15 kJ was evolved

2/ The heat of formation of iron(III) oxide is -826KJ/mol. Calculate the heat of the reaction when a 55.8 g sample of iron is reacted with an excess of oxygen gas.

4Fe(s) + 3O2(g) -----> 2Fe2O3(s)

For 4 mols of Fe we have -826 kJ/mol enthalpy

mols of Fe = 55.8/55.845 = 1.0 mols

So for 1 mols (55.8 g) Fe the enthalpy change would be = -206.5 kJ

3/ the vapor pressure of water at 25*Celcius is 23.8 torr. Determine the mass of glucose( molar mass= 180g/mol) needed to add to 500 grams of water to change the vapor pressure to 23.1 torr.

Moles of water = 500 g / 18.02 g/mol =27.7 mols

23.1 = 23.8 x (mole fraction of water)

Mole fraction of water = 23.1 / 23.8 = 0.971

Let x mols be the amount of glucose added then,

0.971 = moles water / moles water + moles glucose = 27.7 / 27.7 + x

26.9 + ( 0.971x ) = 27.7

x = 0.823 moles glucose

thus,

Mass of glucose needed to be added = 0.823 mol x 180 g/mol = 148 g

4/ Consider the reaction H2(g) + .5O2(g) ----> H2O(I) ∆H*= -286 Kj

the following is true,

A/ The reaction is exothermic

and,

E/ The reaction is exothermic and the enthalpy of the products is less than that of the reactants.