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C(g)+4H(g)-CH4 ∆H=x kj/mol C(graphite,s)+2H2(g)-CH4(g) ∆H=y kj/mol x>y. how.??
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C(g)+4H(g)-CH4 ∆H=x kj/mol C(graphite,s)+2H2(g)-CH4(g) ∆H=y kj/mol x>y...
Explanation of why x>y in the given thermochemical equations:
- Thermochemical equations:
- C(g) + 4H(g) → CH4 ∆H=x kJ/mol
- C(graphite,s) + 2H2(g) → CH4(g) ∆H=y kJ/mol
- Explanation:
- The first equation represents the formation of methane gas (CH4) directly from its elements (C and H2) in their standard states.
- The second equation involves the formation of methane gas from graphite carbon and hydrogen gas.
- Since both equations represent the same reaction but in different forms, their enthalpy changes can be compared.
- The enthalpy change (∆H) for the first equation (x) is greater than the enthalpy change for the second equation (y) because it requires more energy to form methane directly from its gaseous elements than from graphite and hydrogen gas.
- This is due to the stability of the reactants in the second equation, where the elements are in a more stable form (graphite and hydrogen gas) compared to their gaseous form in the first equation.
- Therefore, x is greater than y in this case.
This comparison highlights the differences in enthalpy changes when forming the same product using different starting materials, which can be understood in the context of thermodynamics and chemical reactivity.
Community Answer
C(g)+4H(g)-CH4 ∆H=x kj/mol C(graphite,s)+2H2(g)-CH4(g) ∆H=y kj/mol x>y...
Graphite is standard carbon. As standard state of element's delta H=0 x>y
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