When .0430 mole of titanium is combusted in a bomb calorimeter, the temperature of the calorimeter increases from 30.00degress Celsius to 91.60 degrees Celsius. In a separate experiment, the heat capacity of the calorimeter is measured to be 9.84kJ/K. The heat of reaction for the combustion of a mole of Ti in this calorimeter is ______? The answer is supposed to be -1.52*10^4 Can someone please explain the process of how to get this answer?

asked by @alexs116 • about 2 years ago • Chemistry • 5 pts
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Hey there! In this problem we are given the delta T, Tfinal-Tinitial, which equals to 61.6 Kelvin (remember to convert from Celsius to Kelvin). We also have C of the calorimeter which is measured in kJ/K. So, C=q/deltaT, or 9.84kJ/K= q/61.6 K. From this equation we can solve for q, and it should equal to 606.144 kJ. This is the heat of reaction for the combustion for 0.0430 mole of Ti. We need q for 1 mole.

1 mole Ti *(606.144kJ/0.0430 mole Ti)

moles of Ti cancel out, and we have our heat of reaction for the combustion of 1 mole of Ti which is -1.41*10^4, negative because combustion reaction is exothermic.

Hope that helped, good luck!

answered by @dashab2 • about 2 years ago