Question

Entropy is a state function and its value depends on two or three variable temperature (T), Pressure(P) and volume (V). Entropy change for an ideal gas having number of moles(n) can be determined by the following equation.
`DeltaS=2.303 "nC"_(v)"log"((T_(2))/(T_(1))) + 2.303 "nR log" ((V_(2))/(V_(1)))`
`DeltaS=2.303 "nC"_(P) "log"((T_(2))/(T_(1))) + 2.303 "nR log" ((P_(1))/(P_(2)))`
Since free energy change for a process or a chemical equation is a deciding factor of spontaneity , which can be obtained by using entropy change `(DeltaS)` according to the expression, `DeltaG = DeltaH -TDeltaS` at a temperature `T`
For a reaction `M_(2)O(s) rarr 2M(s) + (1)/(2) O_(2)(g) , DeltaH = 30 KJ//mol` and `DeltaS=0.07 KJ/mole "at" 1` atm. Calculate upto which temperature the reaction would not be spontaneous.
A. `T gt 428.6` K
B. `T gt 300.8` K
C. `T lt 300.8` K
D. `T lt 428.6` K

Answer 1

Correct Answer - D
For reaction to be spontaneous,
`DeltaG lt 0= DeltaH-TDeltaS lt 0`
`implies T gt (DeltaH)/(DeltaS) implies T gt 426.6 K.`
if `T lt 426.6 K` so, reaction is non spontaneous.