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cricket20 [7]
3 years ago
6

In the uncatalyzed reaction N2O4 (g) ⇌ 2 NO2 (g) the pressure of the gases at equilibrium are PN2O4 = 0.377 atm and PNO2 = 1.56

ATM AT 100ºc. What would happen to these pressures if a catalyst were added to the mixture?
Chemistry
1 answer:
larisa86 [58]3 years ago
8 0

Answer:

The pressures will remain at the same value.

Explanation:

A catalyst is a substance that alter the rate of a chemical reaction. It either speeds up the or slows down the rate of a chemical reaction.

While a catalyst affects the rate, it is noteworthy that it has no effect on the equilibrium position of the chemical reaction. A catalyst works by creating an alternative pathway for the reaction to proceed. Most times, it decreases the activation energy needed to kickstart the chemical reaction.

Hence, we know that it has no effect on the equilibrium position. Factors affecting equilibrium position includes, temperature and concentration of reactants and products( pressure in terms of gases).

The reactants and the products here are gaseous, and as such pressure affects the equilibrium position. Now, we have established that the equilibrium position is unaffected. And as such the pressure affecting it does not change.

Thus, we have established that the pressure of the products and reactants are unaffected and as such they remain at their value unaffected.

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It has been suggested that the surface melting of ice plays a role in enabling speed skaters to achieve peak performance. Carry
vesna_86 [32]

Explanation:

Relation between pressure, latent heat of fusion, and change in volume is as follows.

          \frac{dP}{dT} = \frac{L}{T \times \Delta V}

Also, \frac{L}{T} = \Delta S^{fusion}_{m}

where, \Delta V^{fusion}_{m} is the difference in specific volumes.

Hence,    \frac{dP}{dT} = \frac{\Delta S^{fusion}_{m}}{\Delta V^{fusion}_{m}}

As, \Delta S^{fusion}_{m} = \frac{L}{T} = \frac{6010}{273.15} = 22.0 J/mol K

And,   \Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}} ...... (1)

where,    d_{H_{2}O} = density of water

              d_{ice} = density of ice

             M = molar mass of water = 18.02 \times 10^{-3} kg

Therefore, using formula in equation (1) we will calculate the volume of fusion as follows.

        \Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}}

                       = \frac{18.02 \times 10^{-3}}{997} - \frac{18.02 \times 10^{-3}}{920}  

                       = -1.51 \times 10^{-6}        

Therefore, calculate the required pressure as follows.

              \frac{dP}{dT} = \frac{22}{-1.51 \times 10^{-6}}

                              = 1.45 \times 10^{7} Pa/K

or,                           = 145 bar/K

Hence, for change of 1 degree pressure the decrease is 145 bar  and for 4.7 degree change dP = 145 \times 4.7 bar

                              = 681.5 bar

Thus, we can conclude that pressure should be increased by 681.5 bar to cause 4.7 degree change in melting point.

5 0
3 years ago
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