Question

6.45 g of C6H12O6 is burned in a bomb calorimeter containing 950. g of water and the temperature goes from 35.0∘C to 42.3∘C. If the bomb has a heat capacity of 933 J∘C, what is the value for q of the reaction?

Answer 1

Answer:

$q_{rxn}=-488.86\frac{kJ}{mol}$

Explanation:

Hello!

In this case, since this calorimetry problems can by analyzed by figuring out that the heat lost due to the combustion of C6H12O6 is gained by the water and the calorimeter which undergo the mentioned temperature rise, we can write:

$Q_{rxn}=-(m_wC_w+C_c)(T_2-T_1)$

Thus, by plugging in the given data, we obtain:

$Q_{rxn}=-(950.0g*4.184\frac{J}{g\°C} +933\frac{J}{\°C})(42.3\°C-35.0\°C)\\\\Q_{rxn}=-17.5kJ$

Next, we compute the moles of C6H12O6 by using its molar mass (180.18 g/mol) as shown below:

$n=6.45g*\frac{1mol}{180.18g} =0.0358mol$

Thus, the value for q of the reaction turns out:

$q_{rxn}=\frac{Q_{rxn}}{n} \\\\q_{rxn}=\frac{17.5kJ}{0.0358mol} \\\\q_{rxn}=-488.86\frac{kJ}{mol}$

Best regards!

Answer 2

Answer: -35,800 J

Explanation:

To solve this problem, first we an equation to relate the heat transfers involved.

qrxn+qwater+qbomb=0

This is is a consequence of the law of conservation of energy. In other words, this is true because heat can neither be created nor destroyed. Therefore, we solve for qrxn as follows.

qrxn=−[qwater+qbomb]

−[(4.184 Jg∘C)(950. g)(42.3∘C−35.0∘C)+(933 J∘C)(42.3∘C–35.0∘C)]

−(29,016 J+6,811 J)

−35,800 J