If C (diamond) \rightarrow {\C} (graphite) + {\X} {\kJ} {\mol}^{-1} {\C}( diamond )+ {\O}_2( {\g}) \rightarrow {\CO}_2( {\g})+ {\Y} {\kJ} {\mol}^{-1} C (graphite) + {\O}_2( {\g}) \rightarrow {\CO}_2( {\g})+ {\Z} {\kJ} {\mol}^{-1} at constant temperature. Then [29 Jan 2025 Shift 2]

Chemical Thermodynamics Part 2

Section: Chemistry

Question : 5 of 100

Marks: +1, -0

If C (diamond) ⟶C (graphite) +X‌kJ‌mol−1 C( diamond )+O2(g)⟶CO2(g)+Y‌kJ‌mol−1 C (graphite) +O2(g)⟶CO2(g)+Z‌kJ‌mol−1 at constant temperature. Then

[29 Jan 2025 Shift 2]

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