Answer:
C. T>617 K
Explanation:
We are given that
[tex]\Delta H=176KJ/mol[/tex]
[tex]\Delta S=0.285KJ/K\cdot mol[/tex]
We have to find the temperature at which the reaction is spontaneous.
When
[tex]\Delta H>0, \Delta S>0[/tex]
Therefore, the reaction is spontaneous at certain range of temperature.
Option A is not true.
[tex]\Delta G=\Delta H-T\Delta S[/tex]
When [tex]\Delta G[/tex] is negative, then the reaction is spontaneous.
[tex]\Delta G=176-0.285T[/tex]
When T<50
Suppose T=49 K
[tex]\Delta G=176-49\times 0.285>0[/tex]
Therefore, [tex]\Delta G is positive[/tex].Hence, the reaction is not spontaneous.
Option B is wrong.
C.T>617K
Suppose T=618 K
[tex]\Delta G=176-0.285\times 618=-0.13<0[/tex]
Therefore, [tex]\Delta G is negative[/tex].Hence, the reaction is spontaneous.
So, option C is true.
D.T<617 K
Suppose T=616 K
[tex]\Delta G=176-0.285\times 616=0.44>0[/tex]
Therefore, [tex]\Delta G is positive[/tex].Hence, the reaction is not spontaneous.
So, option D is not true.
