**Review**

While working these problems, you may wish to review the section on coffee cup and bomb calorimetry and the laws of thermochemistry.

**Problem**

When a 1.000 g sample of the rocket fuel hydrazine, N_{2}H_{4}, is burned in a bomb calorimeter which contains 1200 g of water, the temperature rises from 24.62°C to 28.16°C. If the C for the bomb is 840 J/°C, calculate

- q
_{reaction}for combustion of a one-gram sample - q
_{reaction}for combustion of one mole of hydrazine in the bomb calorimeter

**Solution**

- For a bomb calorimeter, use this equation:
q

_{reaction}= -(q_{water}+ q_{bomb})q

_{reaction}= -(4.18 J / g·°C x m_{water}x Δt + C x Δt)q

_{reaction}= -(4.18 J / g·°C x m_{water}+ C)Δtwhere q is heat flow, m is mass in grams, and Δt is the temperature change. Plugging in the values given in the problem:

q

_{reaction}= -(4.18 J / g·°C x 1200 g + 840 J/°C)(3.54°C)q

_{reaction}= -20,700 J or -20.7 kJ - We now know that 20.7 kJ of heat is evolved for every gram of hydrazine that is burned. Using the periodic table to get atomic weights, we can calculate that one mole of hydrazine, N
_{2}H_{4}, weight 32.0 g. Therefore, for the combustion of one mole of hydrazine:q

_{reaction}= 32.0 x -20.7 kJ/gq

_{reaction}= -662 kJ

**Answer**

- -20.7 kJ
- -662 kJ

Would you like an example for coffee cup calorimetry? Then return to the previous worked calorimetry problem > Previous Page