The value of the equilibrium constant Kp for the reaction below is 0.747 at 900°C. $$H2(g)+CO2(g)H2O(g)+CO(g) Calculate the value of ΔG° for the reaction. kJ/mol
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314 J/(mol·K)), - \( T \) is the temperature in Kelvin, - \( K \) is the equilibrium constant (Kp in this case). Show more…
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The equilibrium constant (Kp) for the reaction below is 4.40 at 2000. K. H2(g) + CO2(g) ⇌ H2O(g) + CO(g) Calculate ΔG for the reaction. Calculate ΔG for the reaction when the partial pressures are PH2 = 0.20 atm, PCO2 = 0.79 atm, PH2O = 0.64 atm, and PCO = 1.14 atm.
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Calculate the equilibrium constant $K_{\mathrm{p}}$ for the reaction given below if $\Delta G^{\circ}=-10.632 \mathrm{~kJ}$ at $300 \mathrm{~K}$ $$ \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{CO}_{(\mathrm{g})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})} $$
The equilibrium constant $\left(K_{P}\right)$ for the reaction $\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g)$ is 4.40 at 2000 K. (a) Calculate $\Delta G^{\circ}$ for the reaction. (b) Calculate $\Delta G$ for the reaction when the partial pressures are $P_{\mathrm{H}_{2}}=0.25 \mathrm{atm}, P_{\mathrm{CO}_{2}}=0.78 \mathrm{atm}, P_{\mathrm{H}_{2} \mathrm{O}}$ $=0.66 \mathrm{atm},$ and $P_{\mathrm{CO}}=1.20 \mathrm{atm}.$
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