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Goshia [24]
3 years ago
9

The reaction 2 no(g) + cl2(g) → 2 nocl(g) is carried out in a closed vessel. if the partial pressure of no is decreasing at a ra

te of 184 torr/min, what is the rate of change of the total pressure in the vessel? the sign of the answer is important.
Chemistry
1 answer:
BARSIC [14]3 years ago
4 0

Answer:

92.0 torr/min.

Explanation:

  • For the reaction: <em>2NO(g) + Cl₂(g) → 2NOCl(g).</em>

∵ The rate of the reaction = - 1/2(ΔP)NO/Δt = - (ΔP)Cl₂/Δt = 1/2(ΔP)NOCl/Δt.

∵ (ΔP)NO/Δt = - 184 torr/min.

<em>∴ The rate of change of the total pressure in the vessel = - 1/2(ΔP)NO/Δt</em> = (- 1/2)(- 184 torr/min) = <em>92.0 torr/min.</em>

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3 years ago
An ideal gas sealed in a rigid 4.86-L cylinder, initially at pressure Pi=10.90 atm, is cooled until the pressure in the cylinder
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\Delta H=-11897J

Explanation:

Hello,

In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:

\Delta H=\Delta U+V\Delta P

Whereas the change in the internal energy is computed by:

\Delta U=nCv\Delta T

So we compute the initial and final temperatures for one mole of the ideal gas:

T_1= \frac{P_1V}{nR}=\frac{10.90atm*4.86L}{0.082*n}=\frac{646.02K  }{n} \\\\T_2= \frac{P_2V}{nR}=\frac{1.24atm*4.86L}{0.082*n}=\frac{73.49K  }{n}

Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:

\Delta U=1mol*\frac{3}{2} (8.314\frac{J}{mol*K} )*(73.49K-646.02K )=-7140J

Then, the volume-pressure product in Joules:

V\Delta P=4.86L*\frac{1m^3}{1000L} *(1.24atm-10.90atm)*\frac{101325Pa}{1atm} \\\\V\Delta P=-4756.96J

Finally, the change in the enthalpy for the process:

\Delta H=-7140J-4757J\\\\\Delta H=-11897J

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