Calculate the temperature above which the reduction of lead oxide to lead in the following reaction become spontaneous.
`PbO(s) + C(s) rightarrow Pb(s) + CO(g)`
Given `Delta H = 118.4 kJ mol^(-1), Delta S = 200 JK^(-1) mol^(-1)`.

Explanation

To determine the temperature above which the reduction of lead oxide (PbO) to lead (Pb) becomes spontaneous, we need to analyze the Gibbs free energy change (ΔG) for the reaction. The reaction is given as:

\[ \text{PbO(s)} + \text{C(s)} \rightarrow \text{Pb(s)} + \text{CO(g)} \]

### Step 1: Understand the conditions for spontaneity
A reaction is spontaneous when the Gibbs free energy change (ΔG) is negative. The relationship between ΔG, enthalpy change (ΔH), and entropy change (ΔS) is given by the equation:

\[ \Delta G = \Delta H - T \Delta S \]
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