Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M203(s)⇒2M(s)+32O2(g) info given for Gf(kJmol): M203=−6.70 M(s)=0 O2(g)=0 what is the standard change in Gibbs energy for r×n as written in forward direction? (kJ/mol) What is the equilibrium constant (K) of this r×n, as written in forward direction at 298K? What is the equilibrium pressure of O2(g) over M(s) at 298K? (atm)

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Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.  M203(s)⇒2M(s)+32O2(g)  info given for Gf(kJmol):  M203=−6.70  M(s)=0  O2(g)=0  what is the standard change in Gibbs energy for r×n as written in forward direction? (kJ/mol)  What is the equilibrium constant (K) of this r×n, as written in forward direction at 298K?  What is the equilibrium pressure of O2(g) over M(s) at 298K? (atm)

Foreckije · Accepted Answer

Step 1  Given reaction:  M2O3(s)⇒2M(s)+32O2(g)  Given:  M2O3=−6.70kJmol  M(s)=0  O2(s)=0  Calculation:  ΔGr×n0=∑ΔG∏ucts0−∑ΔGrectan⁡ts0  =[(2×0)+(32×0)]−[1×−6.70]  ΔGr×n0=6.70kJmol  Step 2  Given:  T=298K  R=gas constant (8.314JmolK)  ΔG0=6.7kJmol  =6.7kJ×1000JkJ  ΔG0=6700Jmol  Calculation:  ln⁡K=−ΔG0RT  =−6700Jmol8.314JmolK×298K  ln⁡K=−2.704  K=e(−2.704)  K=0.0669  Step 3  M2O3(s)⇒2M(s)+32O2(g)  K=p32O2(g)  pO2(g)=K23  K=0.0669  pO2(g)=(0.0669)23  pO2(g)=0.164at

movingsupplyw1 · Accepted Answer

Step 1  M2O3(s)≤ftharp∞ns2M(s)+32O2(g)  ΔG0=∑npΔGp0−∑nrΔGr0  ⇒2(0)+32(0)−(−6.7)  ΔG0⇒6.7kJmol  Step 2  ΔG0=−RTln⁡(K)  ln⁡(K)=−ΔG0RT  ln⁡(K)=−6.7×1038.314×298=−2.704  K=0.067  Step 3  ΔG0=−RTln⁡(Kp)  Kp=(PO2)32  ΔG0=−RTln⁡[(PO2)32]  ln⁡[(PO2)32]=−ΔG0RT  ln⁡[(PO2)32=−2.704  (PO2)32=e−2.704  (PO2)32=0.067  PO2=0.017atm

Consider the decomposition of a metal oxide to its elements, where M r

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