Question

Part C

If ΔH = -80.0 kJ and ΔS = -0.200 kJ/K , the reaction is spontaneous below a certain temperature. Calculate that temperature.

Express your answer numerically in kelvins.

Answer 1

A reaction is spontaneous when $\Delta$G < 0

But,

$\Delta$G =  $\Delta$H - T $\Delta$S

So,

$\Delta$H - T $\Delta$S < 0

$\Rightarrow$  -80.0 kJ - T K ( -0.2 kJ/K ) < 0

$\Rightarrow$  -80.0 kJ + T x 0.2 kJ < 0

$\Rightarrow$  T x 0.2 kJ < 80.0 kJ

$\Rightarrow$  T < (80.0 kJ) / (0.2 kJ)

$\Rightarrow$  T < 400

So, the temperature is 400 K, below 400 K the reaction is spontaneous.