1.000 g of compound x with molecular formula c₅h₁₀ are burned in a constant - pressure calorimeter…

1.000 g of compound x with molecular formula c₅h₁₀ are burned in a constant - pressure calorimeter containing 10.00 kg of water at 25 °c. the temperature of the water is observed to rise by 1.074 °c. (you may assume all the heat released by the reaction is absorbed by the water, and none by the calorimeter itself.) calculate the standard heat of formation of compound x at 25 °c. be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.

1.000 g of compound x with molecular formula c₅h₁₀ are burned in a constant - pressure calorimeter containing 10.00 kg of water at 25 °c. the temperature of the water is observed to rise by 1.074 °c. (you may assume all the heat released by the reaction is absorbed by the water, and none by the calorimeter itself.) calculate the standard heat of formation of compound x at 25 °c. be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.

Answer

Explanation:

Step1: Calculate heat absorbed by water

The heat - capacity formula is $q = mc\Delta T$. The mass of water $m = 10.00\ kg=10000\ g$, the specific - heat capacity of water $c = 4.184\ J/(g\cdot^{\circ}C)$, and the temperature change $\Delta T=1.074^{\circ}C$. $q = mc\Delta T=10000\ g\times4.184\ J/(g\cdot^{\circ}C)\times1.074^{\circ}C = 44935.16\ J\approx44935\ J$

Step2: Calculate moles of Compound X

The molar mass of $C_5H_{10}$ is $M=(5\times12.01 + 10\times1.01)\ g/mol=(60.05 + 10.1)\ g/mol = 70.15\ g/mol$. The mass of Compound X is $m = 1.000\ g$. The number of moles $n=\frac{m}{M}=\frac{1.000\ g}{70.15\ g/mol}\approx0.01426\ mol$

Step3: Calculate the heat of combustion per mole of Compound X

The heat of combustion per mole $\Delta H_{comb}=\frac{q}{n}=\frac{- 44935\ J}{0.01426\ mol}\approx - 3.15\times10^{6}\ J/mol=-3150\ kJ/mol$

Step4: Write the combustion reaction and use Hess's law

The combustion reaction of $C_5H_{10}$ is $C_5H_{10}(l)+\frac{15}{2}O_2(g)\rightarrow5CO_2(g)+5H_2O(l)$. The standard heats of formation: $\Delta H_f^{\circ}(CO_2(g))=-393.5\ kJ/mol$, $\Delta H_f^{\circ}(H_2O(l))=-285.8\ kJ/mol$, $\Delta H_f^{\circ}(O_2(g)) = 0\ kJ/mol$. By Hess's law, $\Delta H_{comb}=\sum n\Delta H_f^{\circ}(products)-\sum n\Delta H_f^{\circ}(reactants)$. $\Delta H_{comb}=5\Delta H_f^{\circ}(CO_2(g)) + 5\Delta H_f^{\circ}(H_2O(l))-\Delta H_f^{\circ}(C_5H_{10}(l))-\frac{15}{2}\Delta H_f^{\circ}(O_2(g))$. Substitute the known values: $-3150\ kJ/mol=5\times(-393.5\ kJ/mol)+5\times(-285.8\ kJ/mol)-\Delta H_f^{\circ}(C_5H_{10}(l))-\frac{15}{2}\times0\ kJ/mol$. $-3150\ kJ/mol=-1967.5\ kJ/mol - 1429\ kJ/mol-\Delta H_f^{\circ}(C_5H_{10}(l))$. $-3150\ kJ/mol=-3396.5\ kJ/mol-\Delta H_f^{\circ}(C_5H_{10}(l))$. $\Delta H_f^{\circ}(C_5H_{10}(l))=-3396.5\ kJ/mol + 3150\ kJ/mol=-250\ kJ/mol$

Answer:

$-250\ kJ/mol$