For a hypothetical gas-phase reaction, the natural logarithm of the equilibrium constant ( K ) is found to be 40 at T = 300 K and 10 at T = 400 K . Assuming that the standard enthalpy change ( H^ ) and standard entropy change ( S^ ) for the reaction are independent of temperature, at what temperature will the standard Gibbs free energy change ( G^ ) be zero?

Step-by-step solution

The relationship between the equilibrium constant and temperature is given by the van 't Hoff equation: K = - H^ RT + S^ R This indicates that K is a linear function of 1/T . Let y = K and x = 1/T . We are given two points: (x₁, y₁) = (1/300, 40) and (x₂, y₂) = (1/400, 10) . First, we find the slope m of this line: m = y₂ - y₁ x₂ - x₁ = 10 - 40 1 400 - 1 300 = -30 3 - 4 1200 = -30 - 1 1200 = 36000 Using the point-slope form y - y₂ = m(x - x₂) , the equation of the line is: K - 10 = 36000 ( 1 T - 1 400 ) We need to