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    Bhanupriya Peter Kara
    Bhanupriya Peter Kara
    Asked: 3 years agoIn:

    Entropy is a state function and its 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=2.303nC_(v)” “log((T_(2))/(T_(1)))+2.303 nR” “log ((V_(2))/(V_(1)))`
    `DeltaS=2.303nC_(p)” “log((T_(2))/(T_(1)))+2.303 nR” “log ((P_(2))/(P_(1)))`
    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, `Delta G=DeltaH-TDeltaS` at a temperature T.
    For a reaction `M_(2)O(s) to 2M(s)+(1)/(2)O_(2)(g),DeltaH=30` kJ/mol and `DeltaS`=0.07kJ/K-mol at 1 atm. Calculate upto which temperature the reaction would not be spontaneous.
    A. `Tgt428.6K`
    B. `Tgt300.8K`
    C. `Tlt300.8K`
    D. `Tlt428.6K`

    Parabola
    Bhanupriya Peter Kara
    Bhanupriya Peter Kara
    Asked: 3 years agoIn:

    Parabola
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