Question-10 Calculate the work done in following reaction at 27^\circ C.
4SO_2(g)+2O_2(g)\rightarrow4SO_3(g) R=8.314\ JK^{-1}\ mol^{-1}(A) 4988.4\ J
(B) 2494.2\ J
(C) 1247.1\ J
(D) 3741.3\ J
Answer : (A)
Explanation:
The reaction is
4SO_2(g) + 2O_2(g) \rightarrow 4SO_3(g)Δn(g) = 4 − (4 + 2) = −2 mol
Now,
W = −Δn(g)RT
= −(−2) mol × 8.314 J K^{-1} mol^{-1} × 300 K
= 4988.4 J