If for H2(g)+(1/2) O2(g) arrow H2 O(g), Δ H1 is the enthalpy of reaction and for H2(g)+(1/2) O2(g) arrow H2 O(l) Δ H2 is enthalpy of reaction, then

Question

If for H2(g)+(1/2) O2(g) arrow H2 O(g), Δ H1 is the enthalpy of reaction and for H2(g)+(1/2) O2(g) arrow H2 O(l) Δ H2 is enthalpy of reaction, then (A)  Δ H1>Δ H2  (B)  Δ H1=Δ H2  (C)  Δ H1<Δ H2  (D)  Δ H1+Δ H2=0

Tardigrade Admin · Accepted Answer

In the process, H 2(g)+(1/2) O 2(g) longrightarrow H 2 O (l), heat is evolved because state is changing from gaseous to liquid, ie, the process is exothermic. Hence, Δ H1<Δ H2

Q. If for $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g),$
$Δ H_{1}$ is the enthalpy of reaction and for
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$
$Δ H_{2}$ is enthalpy of reaction, then

$Δ H_{1} > Δ H_{2}$

$Δ H_{1} = Δ H_{2}$

$Δ H_{1} < Δ H_{2}$

$Δ H_{1} + Δ H_{2} = 0$

In the process, $H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l)$ , heat is evolved because state is changing from gaseous to liquid, ie, the process is exothermic.
Hence, $< b r / > Δ H_{1} < Δ H_{2}$

Q. If for H2​(g)+21​O2​(g)→H2​O(g), ΔH1​ is the enthalpy of reaction and for H2​(g)+21​O2​(g)→H2​O(l) ΔH2​ is enthalpy of reaction, then

ΔH1​>ΔH2​

ΔH1​=ΔH2​

ΔH1​<ΔH2​

ΔH1​+ΔH2​=0