MHT CET · Chemistry · Thermodynamics (C)
The heat of formation of water is \(260 \mathrm{~kJ}\). How much \(\mathrm{H}_{2} \mathrm{O}\) is decomposed by \(130 \mathrm{~kJ}\) of heat ?
- A \(0.25 \mathrm{~mol}\)
- B \(1 \mathrm{~mol}\)
- C \(0.5 \mathrm{~mol}\)
- D \(2 \mathrm{~mol}\)
Answer & Solution
Correct Answer
(C) \(0.5 \mathrm{~mol}\)
Step-by-step Solution
Detailed explanation
The reaction for the formation of water is as
\(
\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O} ; \quad \Delta H=260 \mathrm{~kJ} \ldots \text { (i) }
\)
On reversing Eq. (i), we get \(\underset{1}{\mathrm{H}_{2} \mathrm{O}} \longrightarrow \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2}\)
\(
\Delta H=-260 \mathrm{~kJ}
\)
\(\because\) By \(260 \mathrm{~kJ}\) heat water decomposed \(=1 \mathrm{~mol}\) \(\therefore 130 \mathrm{k}_{\mathrm{I}} \mathrm{I}\) hont will decomposo water \(=\frac{1 \times 130}{260}\)
\(
=0.5 \mathrm{~mol}
\)
\(
\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O} ; \quad \Delta H=260 \mathrm{~kJ} \ldots \text { (i) }
\)
On reversing Eq. (i), we get \(\underset{1}{\mathrm{H}_{2} \mathrm{O}} \longrightarrow \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2}\)
\(
\Delta H=-260 \mathrm{~kJ}
\)
\(\because\) By \(260 \mathrm{~kJ}\) heat water decomposed \(=1 \mathrm{~mol}\) \(\therefore 130 \mathrm{k}_{\mathrm{I}} \mathrm{I}\) hont will decomposo water \(=\frac{1 \times 130}{260}\)
\(
=0.5 \mathrm{~mol}
\)
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