The standard state Gibbs free energies of formation of $$C(graphite)$$ and $$C(diamond)$$ at T $$=$$ $$298$$ K areΔfG°$$Cgraphite=0$$ kJ mol−$$1$$ΔfG°$$Cdiamonda=2.9$$ kJ mol−$$1$$ The standard state means that the pressure should be $$1$$ bar, and substance should be pure at a given temperature. The conversion of graphite [$$C(graphite)$$ to diamond [C (diamond)] reduces its volume by $$2$$×$$10$$−$$6m3$$ mol−$$1$$. If $$C(graphite)$$ is converted to $$C(diamond)$$ isothermally at T $$=$$ $$298$$ K , the pressure at which $$C(graphite)$$ is in equilibrium with $$C(diamond)$$, is [Useful information: $$1$$ $$J=1$$ kg $$m2s$$−$$2$$ ; $$1$$ $$Pa=1$$ kg m−$$1s$$−$$2$$ ; $$1$$ $$bar=105Pa$$]

Question

The standard state Gibbs free energies of formation of $$C(graphite)$$ and $$C(diamond)$$ at T $$=$$ $$298$$ K areΔfG°$$Cgraphite=0$$ kJ mol−$$1$$ΔfG°$$Cdiamonda=2.9$$  kJ mol−$$1$$ The standard state means that the pressure should be $$1$$ bar, and substance should be pure at a given temperature. The conversion of graphite [$$C(graphite)$$ to diamond [C (diamond)] reduces its volume by $$2$$×$$10$$−$$6m3$$ mol−$$1$$. If $$C(graphite)$$  is converted to $$C(diamond)$$ isothermally at T $$=$$ $$298$$ K , the pressure at which $$C(graphite)$$ is in equilibrium with $$C(diamond)$$, is [Useful information: $$1$$ $$J=1$$  kg $$m2s$$−$$2$$ ; $$1$$ $$Pa=1$$ kg m−$$1s$$−$$2$$ ; $$1$$ $$bar=105Pa$$]

Infinity Learn · Accepted Answer

$$C(g)$$⇌$$C(D)$$ΔGRn $$O=$$Δ$$G(D)O$$−Δ$$G(g)o=2.9$$−$$0=2.9KJ/mol$$Δ$$G0=+P$$Δ$$V2.9kg/m=P2$$×$$10$$−$$6P=1450$$  Bar

Best Courses for You

Detailed Solution

courses

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

download the app

Download the App