Question

For complete combustion of ethanol, the amount of heat produced as measured in a bomb calorimeter is $1364.47kJ$$mo{l}^{-1}$ at $25$${}^{\circ }C$.
$C_2{H}_{5}OH(l)+3O_2(g)\to 2C{O}_2(g)+3H_{2}O(l)$
Assuming ideality the enthalpy of combustion,${△}_{c}H$, for the reaction will be:
[R=8.314J$mo{l}^{-1}$]
(A) $-1460.50kJ$$mo{l}^{-1}$
(B) $-1350.50kJ$$mo{l}^{-1}$
(C) $-1366.95kJ$$mo{l}^{-1}$
(D) $-1361.95kJ$$mo{l}^{-1}$

Answer 1

Hint: In a bomb calorimeter, volume is always constant and heat produced is equal to the internal energy. Now, apply the internal energy and enthalpy formula and solve the problem.

Complete step by step solution:
Bomb calorimeter is a thick steel container which is used to determine the energy changes associated with the chemical and physical process i.e. the energy which is contained in the substance by calculating the heat generated during the combustion of the substance which is also known as enthalpy of combustion( means heat released when one mole of the substance is completely burned) and in the bomb calorimeter the volume is constant.
Now, in the statement it is given that:
The amount of heat produced in bomb calorimeter is $1364.47kJ$$mo{l}^{-1}$. since we know that in bomb calorimeter the volume is constant so the heat produced is directly proportional to the change in the internal energy $\mathrm{\Delta }U$ i.e.
$\mathrm{\Delta }U$ $\alpha$$\mathrm{\Delta }q$
$\mathrm{\Delta }U$= c$mo{l}^{-1}$
Since heat is released so, $\mathrm{\Delta }U$ is taken as negative,
$\mathrm{\Delta }U$= -1364.47kJ$mo{l}^{-1}$
Now, we have to calculate the enthalpy of combustion i.e. ${△}_{c}H$
And we know that the $\mathrm{\Delta }U$and${△}_{c}H$ are related to each other by the equation:
${△}_{c}H$= $\mathrm{\Delta }U$+$\mathrm{\Delta }{n}_g$RT
Here, ${△}_{c}H$ is the enthalpy of combustion, $\mathrm{\Delta }U$ is the change in internal energy , $\mathrm{\Delta }{n}_g$ is the no moles of gaseous reactants minus the no of moles gaseous products’ is the gas constant whose value is $R=8.314J$$mo{l}^{-1}$ and $T$ is the temperature.
Temperature, $T$ has been in the question and $\mathrm{\Delta }U$ is been given.
Now, we will calculate the no of moles:
$C_2{H}_{5}OH(l)+3O_2(g)\to 2C{O}_2(g)+3H_{2}O(l)$
$\mathrm{\Delta }{n}_g$= no of moles of gaseous reactants – no of moles of gaseous products
$=1$
Temperature=$25$${}^{\circ }C$ (given)
Covert, it into kelvin, we get
$T=$ $25+273$
$=298K$
Now, putting all the values in equation (1), we get,
${△}_{c}H$=-1364.47 + ${\displaystyle \frac{-1\times 8.314\times 298}{1000}}$ (here, the value of R must be converted into KJ from J)
$=$$-1364.47-2.4776$
$=-1366.95KJ$$mo{l}^{-1}$

Hence, option (C) is correct.

Note: whenever the heat is released, always put the negative sign because it involves exothermic reaction and always remember that temperature is always taken in kelvin and the value of R can be taken in both J and KJ depending upon the options available to us.