For which of the following reaction triangle H-triangle E-2RTP- Cl - 5 -left- g right- rightarrow PC- l - 3 -left- g right- - Cl - 2 -left- g right- - N - 2 -left- g right- - O - 2 -left- g right- rightarrow 2NOleft- g right- - N - 2 -left- g right- -3- H - 2 -left- g right- rightarrow 2- NH - 3 -left- g right- - NH - 4 -HSleft- s right- rightarrow - NH - 3 -left- g right- - H - 2 -Sleft- g right-

Question

Toppr Admin · Accepted Answer

-x394-H-x394-E-x394-ngasRT-xA0-Comparing with -x394-H-x394-E-2RT-xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -x394-ngas-21-x2192-x394-ngas-2-x2212-1-x2212-3-x2212-2-2-x2192-x394-ngas-2-x2212-1-x2212-1-03-x2192-x394-ngas-1-1-x2212-0-24-x2192-x394-ngas-1-1-x2212-1-1

For which of the following reaction $△ H = △ E + 2 R T$
$P {C l}_{5} (g) \to P C l_{3} (g) + {C l}_{2} (g)$
$N_{2} (g) + O_{2} (g) \to 2 N O (g)$
$N_{2} (g) + 3 H_{2} (g) \to 2 {N H}_{3} (g)$
${N H}_{4} H S (s) \to {N H}_{3} (g) + H_{2} S (g)$

$Δ H = Δ E + Δ n_{g a s} R T$
Comparing with $Δ H = Δ E + 2 R T$
$Δ n_{g a s} = 2$
$1 \to Δ n_{g a s} = 2 - 1 - 3 = (- 2)$
$2 \to Δ n_{g a s} = 2 - 1 - 1 = 0$
$3 \to Δ n_{g a s} = 1 + 1 - 0 = 2$
$4 \to Δ n_{g a s} = 1 + 1 - 1 = 1$

For which of the following reaction △H=△E+2RTPCl5(g)→PCl3(g)+Cl2(g)N2(g)+O2(g)→2NO(g)N2(g)+3H2(g)→2NH3(g)NH4HS(s)→NH3(g)+H2S(g)

ΔH=ΔE+ΔngasRT Comparing with ΔH=ΔE+2RT Δngas=21→Δngas=2−1−3=(−2)2→Δngas=2−1−1=03→Δngas=1+1−0=24→Δngas=1+1−1=1

For which of the following reaction $△ H = △ E + 2 R T$
$P {C l}_{5} (g) \to P C l_{3} (g) + {C l}_{2} (g)$
$N_{2} (g) + O_{2} (g) \to 2 N O (g)$
$N_{2} (g) + 3 H_{2} (g) \to 2 {N H}_{3} (g)$
${N H}_{4} H S (s) \to {N H}_{3} (g) + H_{2} S (g)$

$Δ H = Δ E + Δ n_{g a s} R T$
Comparing with $Δ H = Δ E + 2 R T$
$Δ n_{g a s} = 2$
$1 \to Δ n_{g a s} = 2 - 1 - 3 = (- 2)$
$2 \to Δ n_{g a s} = 2 - 1 - 1 = 0$
$3 \to Δ n_{g a s} = 1 + 1 - 0 = 2$
$4 \to Δ n_{g a s} = 1 + 1 - 1 = 1$