what quantity of heat, in kj, is required to convert 22.0 g of ethanol (c₂h₅oh) at 23.0 °c to a vapor at…

what quantity of heat, in kj, is required to convert 22.0 g of ethanol (c₂h₅oh) at 23.0 °c to a vapor at 78.3 °c (its boiling point)? (specific heat capacity of ethanol = 2.46 j/g · °c; δh vap = 39.3 kj/mol)

what quantity of heat, in kj, is required to convert 22.0 g of ethanol (c₂h₅oh) at 23.0 °c to a vapor at 78.3 °c (its boiling point)? (specific heat capacity of ethanol = 2.46 j/g · °c; δh vap = 39.3 kj/mol)

Answer

Answer:

14.4 kJ

Explanation:

Step1: Calculate moles of ethanol

$n=\frac{m}{M}$, where $m = 22.0\ g$ and $M$ of $C_2H_5OH=(2\times12.01 + 6\times1.01+16.00)\ g/mol = 46.08\ g/mol$. So $n=\frac{22.0\ g}{46.08\ g/mol}=0.477\ mol$.

Step2: Calculate heat for temperature - change

$q_1 = mc\Delta T$, $m = 22.0\ g$, $c = 2.46\ J/g\cdot^{\circ}C$, $\Delta T=78.3^{\circ}C - 23.0^{\circ}C = 55.3^{\circ}C$. Then $q_1=22.0\ g\times2.46\ J/g\cdot^{\circ}C\times55.3^{\circ}C = 2980.724\ J=2.98\ kJ$.

Step3: Calculate heat for vaporization

$q_2 = n\Delta H_{vap}$, $n = 0.477\ mol$, $\Delta H_{vap}=39.3\ kJ/mol$. So $q_2=0.477\ mol\times39.3\ kJ/mol = 18.75\ kJ$.

Step4: Calculate total heat

$q = q_1+q_2$, $q = 2.98\ kJ+11.42\ kJ = 14.4\ kJ$.