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At 1813C the equilibrium constant for the reaction: 2IBr(g)I2(g)+Br2(g) is Kp=0.725. If the initial pressure of IBr is 0.00294atm, what are the equilibrium partial pressures

At 1813C the equilibrium constant for the reaction: 2IBr(g)I2(g)+Br2(g) is Kp=0.725. If the initial pressure of IBr is 0.00294atm, what are the equilibrium partial pressures of IBr,I2, and Br2 ? p(IBr)= p(T2)= p(Br2)=

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