The enthalpies of formation of N 2 O and NO are 30 and 90 kJ mol -1 respectively. The enthalpy of the reaction is x × 102 kJ 2 N 2 O ( g )+ O 2( g ) longrightarrow 4 NO ( g ) the value of x is ?

Question

The enthalpies of formation of N 2 O and NO are 30 and 90 kJ mol -1 respectively. The enthalpy of the reaction is x × 102 kJ 2 N 2 O ( g )+ O 2( g ) longrightarrow 4 NO ( g ) the value of x is ? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

N 2( g )+(1/2) O 2( g ) longrightarrow N 2 O ( g ) Δ H =30 kJ ldots (i) (1/2) N2(g)+(1/2) O2(g) longrightarrow N O(g) Δ H =90 kJ ...(ii) By eq [4 × (ii) -2 × (i) ] 2 N 2 O + O 2 longrightarrow 4 NO Δ H =300 kJ 3 × 102 kJ ∴ x =3

Q. The enthalpies of formation of $N_{2} O$ and $NO$ are $30$ and $90 k J m o l^{- 1}$ respectively. The enthalpy of the reaction is $x \times 1 0^{2} k J$
$2 N_{2} O (g) + O_{2} (g) ⟶ 4 NO (g)$
the value of $x$ is ?

$N_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ N_{2} O (g) Δ H = 30 k J \dots$ (i)
$\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ NO (g) Δ H = 90 k J$ ...(ii)
By eq $[4 \times$ (ii) $- 2 \times$ (i) $]$
$2 N_{2} O + O_{2} ⟶ 4 NO Δ H = 300 k J$
$3 \times 1 0^{2} k J ∴ x = 3$

Q. The enthalpies of formation of N2​O and NO are 30 and 90kJmol−1 respectively. The enthalpy of the reaction is x×102kJ 2N2​O(g)+O2​(g)⟶4NO(g) the value of x is ?

N2​(g)+21​O2​(g)⟶N2​O(g)ΔH=30kJ… (i) 21​N2​(g)+21​O2​(g)⟶NO(g)ΔH=90kJ...(ii) By eq [4× (ii) −2× (i) ] 2N2​O+O2​⟶4NOΔH=300kJ 3×102kJ∴x=3

Q. The enthalpies of formation of $N_{2} O$ and $NO$ are $30$ and $90 k J m o l^{- 1}$ respectively. The enthalpy of the reaction is $x \times 1 0^{2} k J$
$2 N_{2} O (g) + O_{2} (g) ⟶ 4 NO (g)$
the value of $x$ is ?

$N_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ N_{2} O (g) Δ H = 30 k J \dots$ (i)
$\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ NO (g) Δ H = 90 k J$ ...(ii)
By eq $[4 \times$ (ii) $- 2 \times$ (i) $]$
$2 N_{2} O + O_{2} ⟶ 4 NO Δ H = 300 k J$
$3 \times 1 0^{2} k J ∴ x = 3$