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The standard enthalpy and entropy changes for a reversible reaction are given below:
at $300 K, Δ H^{0} = - 40 {kJ mol}^{-1}; Δ S^{0} = - 4 \times 10^{- 2} {kJ mol}^{-1}$
at $1200 K, Δ H^{0} = - 33 {kJ mol}^{-1}; Δ S^{0} = - 3 \times 10^{- 2} {kJ mol}^{-1}$ . (consider R=8 joules per kelvin (K) per mole.) Identify correct statements.

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The reaction is reactant favorable at 1200 K

When temperature is raised from 300K to 1200K, the reaction proceeds in the forward direction

The reaction is product favorable at 1200 K

The reaction is product favorable at 300 K.

answer is A, C.

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Detailed Solution

At $300 K, Δ G^{0} = Δ H^{0} - T Δ S^{0} = - 28.44 kJ/mole$ proceeds in forward direction
At $1200 K, Δ G^{0} = + 2.59 kJ/mole$ proceeds in reverse direction

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