If for H2g - 12 O2g - H2Og- H1 is the enthalpy of reaction and for H2g - 12O2g - H2Ol- H2 is enthalpy of reaction then-

The correct option is C H_1 In the process, H_2(g) + 12O_2(g)→H_2O(l) ; heat is evolved because the state is changing from gaseous to liquid, i.e., th ...

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Standard XII

Chemistry

Hess' Law

If for H2g ...

Question

If for $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g); △ H_{1}$ is the enthalpy of reaction and for $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); △ H_{2}$ is enthalpy of reaction then:

△ H_{1} > △ H_{2}

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△ H_{1} = △ H_{2}

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△ H_{1} < △ H_{2}

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△ H_{1} + △ H_{2} = 0

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Solution

The correct option is C $△ H_{1} < △ H_{2}$
In the process, $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$ ; heat is evolved because the state is changing from gaseous to liquid, i.e., the process is more exothermic because of gas-phase changing to the liquid phase.

Hence, $△ H_{1} < △ H_{2}$ .

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If for H2(g)+12O2(g)→H2O(g);△H1 is the enthalpy of reaction and for H2(g)+12O2(g)→H2O(l);△H2 is enthalpy of reaction then:

The correct option is C △H1<△H2In the process, H2(g)+12O2(g)→H2O(l); heat is evolved because the state is changing from gaseous to liquid, i.e., the process is more exothermic because of gas-phase changing to the liquid phase. Hence, △H1<△H2.

If for $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g); △ H_{1}$ is the enthalpy of reaction and for $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); △ H_{2}$ is enthalpy of reaction then: