COMEDK · Chemistry · 6. Thermodynamics (C)
The standard enthalpies of formation of CH\(_4\)(g), CO\(_2\)(g) and H\(_2\)O(l) are \(-74.8\) kJ mol\(^{-1}\), \(-393.5\) kJ mol\(^{-1}\) and \(-285.8\) kJ mol\(^{-1}\) respectively. Then the enthalpy change for the given reaction in kJ mol\(^{-1}\) will be:
\(2\text{CH}_4(g) + 4\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 4\text{H}_2\text{O}(l)\)
- A \(-890.3\)
- B \(+1780.6\)
- C \(-1780.6\)
- D \(+890.3\)
Answer & Solution
Correct Answer
(C) \(-1780.6\)
Step-by-step Solution
Detailed explanation
The enthalpy of reaction is calculated using the standard enthalpies of formation of the products and reactants. \(\Delta_{r}H^{\circ} = \sum \Delta_{f}H^{\circ}(\text{products}) - \sum \Delta_{f}H^{\circ}(\text{reactants})\) For the given reaction:…
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