H2(g) + (1/2)O2(g) → H2O(l) H2O(l) → H2O(g) ; Δ H = x4 Given, EH - H = x1 EO = O = x2 EO - H = x3 Δ HF of H2O vapour is

Question

H2(g) + (1/2)O2(g) → H2O(l) H2O(l) → H2O(g) ; Δ H = x4 Given, EH - H = x1 EO = O = x2 EO - H = x3 Δ HF of H2O vapour is (A) x1+(x2/2)-x3+x4 (B) 2 x 3- x 1-( x 2/2)- x 4 (C) x1+(x2/2)-2 x3-x4 (D) x1+(x2/2)-2 x3+x4

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/9d4904bfffb287857251ea85f2585684-.png" /> Now, for H 2( g )+(1/2) O 2( g ) longrightarrow H 2 O (ℓ) ; Δ H Δ H =( B . D . E ) text reactants -( B . D . E ) text products = x 1+(1/2) x 2-2 x 3 Δ H f =Δ H + x 4 = x 1+(1/2) x 2-2 x 3+ x 4

Q. $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$
$H_{2} O_{(l)} \to H_{2} O_{(g)}; Δ H = x_{4}$
Given, $E_{H - H} = x_{1}$
$E_{O = O} = x_{2}$
$E_{O - H} = x_{3}$
$Δ H_{F}$ of $H_{2} O$ vapour is

$x_{1} + \frac{x _{2}}{2} - x_{3} + x_{4}$

$2 x_{3} - x_{1} - \frac{x _{2}}{2} - x_{4}$

$x_{1} + \frac{x _{2}}{2} - 2 x_{3} - x_{4}$

$x_{1} + \frac{x _{2}}{2} - 2 x_{3} + x_{4}$

Now, for $H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (ℓ); Δ H$
$Δ H = (B . D . E)_{reactants} - (B . D . E)_{products}$
$= x_{1} + \frac{1}{2} x_{2} - 2 x_{3}$
$Δ H_{f} = Δ H + x_{4}$
$= x_{1} + \frac{1}{2} x_{2} - 2 x_{3} + x_{4}$

Q. H2​(g)+21​O2​(g)→H2​O(l) H2​O(l)​→H2​O(g)​;ΔH=x4​ Given, EH−H​=x1​ EO=O​=x2​ EO−H​=x3​ ΔHF​ of H2​O vapour is

x1​+2x2​​−x3​+x4​

2x3​−x1​−2x2​​−x4​

x1​+2x2​​−2x3​−x4​

x1​+2x2​​−2x3​+x4​