Enthalpy of formation of NH3 is - X kJ and ΔHH-H, ΔHN-H are respectively Y kJ mol-1 and Z kJ mol-1. The value of ΔHN ≡ N is

Question

Enthalpy of formation of NH3 is - X kJ and ΔHH-H, ΔHN-H are respectively Y kJ mol-1 and Z kJ mol-1. The value of ΔHN ≡ N is (A) Y-6 Z+(X/3) (B) -3 Y+6 Z-2 X (C) 3 Y+6 Z+X (D) Y+6 X+Z

Tardigrade Admin · Accepted Answer

(1/2) N 2( g )+(3/2) H 2( g ) longrightarrow NH 3( g ) ; Δ H f =- x kJ Δ H f =Δ H =( B .D . E ) text reactants -( B .D .E ) text products =(1/2) Δ H N 2+(3/2) Δ H H 2-3 Δ N - H =(Δ HN ≡ N/2)+(3/2) X Y-3 X Z We have, -X=(Δ HN ≡ N/2)+(3 Y/2)-3 Z or, Δ H N ≡ N =(3 Z-(3 Y/2)-X) × 2 =6 Z-3 Y-2 X

Q. Enthalpy of formation of $N H_{3}$ is $- X k J$ and $Δ H_{H - H}, Δ H_{N - H}$ are respectively $Y k J m o l^{- 1}$ and $Z k J m o l^{- 1}$ . The value of $Δ H_{N \equiv N}$ is

$Y - 6 Z + \frac{X}{3}$

$- 3 Y + 6 Z - 2 X$

$3 Y + 6 Z + X$

$Y + 6 X + Z$

Q. Enthalpy of formation of NH3​ is −XkJ and ΔHH−H​,ΔHN−H​ are respectively YkJmol−1 and ZkJmol−1. The value of ΔHN≡N​ is

Y−6Z+3X​

−3Y+6Z−2X

3Y+6Z+X

Y+6X+Z

21​N2​(g)+23​H2​(g)⟶NH3​(g);ΔHf​=−xkJ ΔHf​=ΔH=(B.D.E)reactants ​−(B.D.E)products ​ =21​ΔHN2​​+23​ΔHH2​​−3ΔN−H​ =2ΔHN≡N​​+23​XY−3XZ We have, −X=2ΔHN≡N​​+23Y​−3Z or, ΔHN≡N​=(3Z−23Y​−X)×2 =6Z−3Y−2X

Q. Enthalpy of formation of $N H_{3}$ is $- X k J$ and $Δ H_{H - H}, Δ H_{N - H}$ are respectively $Y k J m o l^{- 1}$ and $Z k J m o l^{- 1}$ . The value of $Δ H_{N \equiv N}$ is

$Y - 6 Z + \frac{X}{3}$

$- 3 Y + 6 Z - 2 X$

$3 Y + 6 Z + X$

$Y + 6 X + Z$