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For the reaction A + 2B ←→ C, what is the value of KC if the rate law is given by –rA = k(CA2/3/CB – KCCC)? Given that initial concentration of A is equal to that of B = 5M and the equilibrium conversion of B is found to be 0.3. k = 100.

For the reaction A + 2B ←→ C, what is the value of KC if the rate law is given by –rA = k(CA2/3/CB – KCCC)? Given that initial concentration of A is equal to that of B = 5M and the equilibrium conversion of B is found to be 0.3. k = 100. Correct Answer Kc = 1

XA = CBoaXB/CAob = 0.15 CA = CAo(1 – XA) CB = CAo(1 – 2XA) CC = CAoXA At equilibrium, take –rA equal to 0 and solve for KC.

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