Chemistry - Thermodynamics - I Question with Solution | TestHub

The reaction is $3A(g)+2B(g)⟶4C(g)$.

Initial moles: $A=5$, $B=4$.

To find the limiting reactant:

For A: $5/3=1.67$

For B: $4/2=2$

Since $1.67<2$, A is the limiting reactant.

Moles of A consumed = 5.

Moles of B consumed = $(2/3)\times 5=10/3$.

Moles of C formed = $(4/3)\times 5=20/3$.

Change in moles of gas, $\Delta n_g=(molesofgaseousproducts)-(molesofgaseousreactantsconsumed)$

$\Delta n_g=(20/3)-(5+10/3)=(20/3)-(15/3+10/3)=(20/3)-(25/3)=-5/3$ mol.

Work done, $W=-\Delta n_{g}RT$.

$W=-(-5/3)\times 2cal/mol-K\times 300K$

$W=(5/3)\times 2\times 300=5\times 2\times 100=1000cal$.