H-2-g- - Cl-2 -g- - 2HCl -g-Delta-H-298K- - -22-06 kcal- For this reaction- Delta-U- is equal to-22-06 - 2 times 298 times 4 kcal-22-06 - 2 times 298 kcal-22-06 - 2times 10-3- times 298 times 2 kcal-22-06 kcal

Question

Toppr Admin · Accepted Answer

Enthalpy of reaction-x394-H-x394-E-ngR-xA0-timesT-x394-E-x394-H-x2212-ngR-xD7-T-x394-E-x2212-22-06-x2212-0-xD7-RT-x2212-22-06Kcal

$H_{2} (g) + C l_{2} (g) = 2 H C l (g)$ ;
$Δ H$ (298K) $= - 22.06$ kcal. For this reaction, $Δ U$ is equal to:
$- 22.06 - 2 \times 298 \times 4 k c a l$
$- 22.06 + 2 \times 298 k c a l$
$- 22.06 + 2 \times 10^{- 3} \times 298 \times 2 k c a l$
$- 22.06 k c a l$

Enthalpy of reaction=
$Δ H = Δ E + n_{g} R t i m e s T$
$Δ E = Δ H - n_{g} R \times T$
$Δ E = - 22.06 - 0 \times R T = - 22.06 K c a l$

H2(g)+Cl2(g)=2HCl(g);ΔH(298K) =−22.06kcal. For this reaction, ΔU is equal to:−22.06−2×298×4 kcal−22.06+2×298 kcal−22.06+2×10−3×298×2kcal−22.06 kcal

Enthalpy of reaction=ΔH=ΔE+ngR timesTΔE=ΔH−ngR×TΔE=−22.06−0×RT=−22.06Kcal

$H_{2} (g) + C l_{2} (g) = 2 H C l (g)$ ;
$Δ H$ (298K) $= - 22.06$ kcal. For this reaction, $Δ U$ is equal to:
$- 22.06 - 2 \times 298 \times 4 k c a l$
$- 22.06 + 2 \times 298 k c a l$
$- 22.06 + 2 \times 10^{- 3} \times 298 \times 2 k c a l$
$- 22.06 k c a l$

Enthalpy of reaction=
$Δ H = Δ E + n_{g} R t i m e s T$
$Δ E = Δ H - n_{g} R \times T$
$Δ E = - 22.06 - 0 \times R T = - 22.06 K c a l$