For the reaction: CO ( g )+(1/2) O 2( g ) arrow CO 2( g ) Δ H°=-67650 cal text at 25° C Calculate Δ H° at 100° C, given that the required molar heat capacities are as follows: CP( CO, gas )=6.97 cal / ° C CP( CO 2., gas )=8.47 cal / ° C CP( O 2., gas )=7.00 cal / ° C

Question

For the reaction: CO ( g )+(1/2) O 2( g ) arrow CO 2( g ) Δ H°=-67650 cal text at 25° C Calculate Δ H° at 100° C, given that the required molar heat capacities are as follows: CP( CO, gas )=6.97 cal / ° C CP( CO 2., gas )=8.47 cal / ° C CP( O 2., gas )=7.00 cal / ° C (A) -54.6 cal (B) -67650.4 cal (C) -67684.4 cal (D) -67762.5 cal

Tardigrade Admin · Accepted Answer

Δ H100°-=Δ H25° C °=(Cp2-Cp2) Δ T ⇒ Δ H100° =75(8.97-6.97-(7/2))+Δ H25° C ° =(-112.50-67650) cal =-67762.5 cal

$- 54.6 c a l$

$- 67650.4 c a l$

$- 67684.4 c a l$

$- 67762.5 c a l$

$Δ H_{100}^{\circ} - = Δ H_{2 5^{\circ} C}^{\circ} = (C_{p_{2}} - C_{p_{2}}) Δ T$
$\Rightarrow Δ H_{100}^{\circ} = 75 (8.97 - 6.97 - \frac{7}{2}) + Δ H_{2 5^{\circ} C}^{\circ}$
$= (- 112.50 - 67650) c a l = - 67762.5 c a l$

Q. For the reaction: CO(g)+21​O2​(g)→CO2​(g) ΔH∘=−67650cal at 25∘C Calculate ΔH∘ at 100∘C, given that the required molar heat capacities are as follows: CP​(CO, gas )=6.97cal/∘C CP​(CO2​, gas )=8.47cal/∘C CP​(O2​, gas )=7.00cal/∘C

−54.6cal

−67650.4cal

−67684.4cal

−67762.5cal

ΔH100∘​−=ΔH25∘C∘​=(Cp2​​−Cp2​​)ΔT ⇒ΔH100∘​=75(8.97−6.97−27​)+ΔH25∘C∘​ =(−112.50−67650)cal=−67762.5cal

$- 54.6 c a l$

$- 67650.4 c a l$

$- 67684.4 c a l$

$- 67762.5 c a l$

$Δ H_{100}^{\circ} - = Δ H_{2 5^{\circ} C}^{\circ} = (C_{p_{2}} - C_{p_{2}}) Δ T$
$\Rightarrow Δ H_{100}^{\circ} = 75 (8.97 - 6.97 - \frac{7}{2}) + Δ H_{2 5^{\circ} C}^{\circ}$
$= (- 112.50 - 67650) c a l = - 67762.5 c a l$