Question

10) pts) For the unbalanced equation: C6H6 (1) + O2(g) → CO2(g) + H20 (1) AH = - 6535 kJ. What is AH when 45.67 g of CO2 is produced? Cottotit O2(g) 02 (g) thout Solo 2 11) (12 pts) A scientist mixed 25.00 mL of 2.00 M KOH with 25.00 mL of 2.00 M HBr. The temperature of the mixed solution rose from 22.7 °C to 31.9°C. Assume that the volumes are additive, and the density of the solution is 1.00 g/mL, and that the specific heat of the solution is 4.184 J/g °C. What is the AH of the reaction in kJ? q = S.H. xm x AT Pis constant.

Answer 1

10) The balanced reaction is:

$C_{6}H_{6,(l)}+\frac{15}{2}O_{2,(g)}\rightarrow 6CO_{2,(g)}+3H_{2}O_{(l)}$

I've written it like this, using fractional coefficients, since it is the most useful way for what follows. The given ∆H is a molar value, which means that it represent the heat generated by the reaction of 1 mole of C₆H₆, which is exactly what we have in our reaction. As you can see, when 1 mole of C₆H₆ reacts, 6 moles of carbon dioxide are generated. These 6 moles are equal to a mass of:

$m=n\cdot m_{molar}=6moles\cdot 44\frac{g}{mole}=264g$

This is the mass of carbon dioxide that is generated when 1 mole of C₆H₆ reacts, which is equivalent to say that this is the mass generated when the heat released is -6535 kJ. Knowing this, we can calculate the heat released when 45.67 g of carbon dioxide are produced using cross multiplication:

$\Delta H_{reaction}=45.67g\cdot \frac{-6535kJ}{264g}=-1131kJ$

11) The reaction taking place here is:

$KOH+HBr\rightarrow KBr+H_{2}O$

And the amount to reactants participating is the same (same volume and concentration). The number of moles of both is:

$n=M\cdot V(L)=2.00M\cdot 0.02500L=0.05moles$

And the heat generated from the reaction can be calculated using the given expression, taking into account that the final mass of the system is 50 g (25 ml + 25 mL, with a density of 1.00 g/mL) and that the ∆T is: (31.9-22.7)°C = 9.2°C:

$q=S.H.\cdot m\cdot \Delta T=4.184\frac{J}{g\cdot ^{o}C}\cdot 50.00g\cdot 9.2^{o}C=1925J$

This is the heat received by the solution, which has the same numerical value as the heat generated by the reaction, but with the opposed sign. The heat "given" by the reaction is thus -1925 J.
Ther'e one more thing to bear in mind, and it is the fact that if we want the ∆H value per mole, we need to divide the heat released by OUR reaction, by the number of moles that participated:

$\Delta H=\frac{-1925J}{0.05moles}=-38493J/mol=-38.5kJ/mol$