Question

For the reaction of one-mole zinc dust with one mole sulfuric acid in a bomb calorimeter, $\Delta U$and $w$ correspond to:
Options are-
(A) $\Delta U$= 0, $w$ = 0
(B) $\Delta U$< 0, $w$ < 0
(C) $\Delta U$> 0, $w$ = 0
(D) $\Delta U$> 0, $w$ > 0

Answer 1

Hint: In a bomb calorimeter the heat supplied (Q) is zero. The change in internal energy of the system ( $\Delta U$) can be calculated by applying the first law of thermodynamics. The work done can be calculated by the below mentioned formula:
$w\text{ = - }\Delta {{\text{n}}_{g}}\text{RT}$
Where,
$\Delta {{\text{n}}_{g}}$stands for difference of number of moles of gaseous molecules in products and reactants,
R stands for universal gas constant,
T is the temperature at which the reaction is carried out.

Complete answer:
Bomb calorimeter is a type of constant-volume calorimeter used to measure the heat of combustion of a reaction. Bomb calorimeters withstand large pressure within the calorimeter as the reaction has to be measured.
The first law of thermodynamics is a version of the law of conservation of energy, applied for thermodynamic processes, distinguishing the two kinds of transfer of energy, namely heat and thermodynamic work, and relating them to a function of the body's state, called Internal energy.
Kinetic energy is defined as the sum of potential and kinetic energy associated with the system.
The equation for first law of thermodynamics is given below:
$\Delta U$ = Q + W
Where,
$\Delta U$is the change in internal energy
Q is the amount of heat supplied to the container(here 0),
W is the amount of thermodynamic work done.
Since the value of Q is zero in bomb calorimeters, $\Delta U$is equal to the thermodynamic work done.
The chemical reaction is given below:
$\text{Z}{{\text{n}}_{(s)}}\text{ + }{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}\text{ }\to \text{ ZnS}{{\text{O}}_{4(aq.)}}\text{ + }{{\text{H}}_{2(g)}}$
For the above reaction,
$\Delta {{\text{n}}_{g}}$is equal to 1.
Substituting the value of $\Delta {{\text{n}}_{g}}$in the formula to calculate work done we get:
$w\text{ = - RT}$
Hence, we can conclude that the work done is negative as R and T are always positive.
Since work done is negative, $\Delta U$is negative as well.

Therefore, the correct answer is option (B).

Note:
Before applying the first law of thermodynamics to a reaction, we make the following assumptions:
-Mass flows in or out of the system along one boundary of the system only.
-The rate of flow of mass into the system is taken as negative and mass flow out of the system is taken as positive.
-Mass can carry the internal energy into or out of the thermodynamic system.