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Show that R=8.314 Jmol^(-1)K^(-1)...

Show that R=8.314 `Jmol^(-1)K^(-1)`

Text Solution

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We know that `(PV)/T=R`
At normal temperature `P_0=1atm=1.013xx10^5Nm^(-2),V_0=0.0224m^3` and `T_0=273K`
Hence `R=((1.013xx10^5)(0.0224))/(273)=8.314Jmol^(-1)K^(-1)`
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In general it is observed that the rate of a chemical reaction doubles with every 10 degree rise in temperature. If the generalization holds good for the reaction in the temperature range 295 K to 305 K, what would be the value of activation energy for this reaction ? [R = 8.314 mol^(-1)JK^(-1)]

What is the value of the equilibrium constant for the following reaction at 400 K ? 2NOCl(g) hArr 2NO(g) + Cl_(2)(g) Delta H^(@) = 77.5 kJ mol^(-1), R = 8.3124 J mol^(-1) K^(-1), Delta S = 135 J K^(-1) mol^(-1) .

Knowledge Check

  • Two reaction R_1 and R_2 have identical pr-exponential factor. Activation energy of R_1 exceeds that of R_2 by 10 kJ mol^(-1) . If k_1 and k_2 re rate constant for reactions R_1 and R_2 respectively at 300 K, then ln (k_2//k_1) is equal to : (R=8cdot314 J mol ^(-1) K^(-1)

    A
    8
    B
    6
    C
    12
    D
    4

  • What is the activation energy for a reaction if the rate is doubled when the temperature is raised from 20^@C to 35^@C (R= 8 cdot 314 J mol^(-1) K^(-1)) .

    A
    `15 cdot 1 kJ mol^(-1)
    B
    342 kJ mol^(-1)`
    C
    `269 k Jmol^(-1)`
    D
    `34 cdot 7 kJ mol^(-1)

  • Assuming that water vapour is an ideal gas, the internal energy changes (Delta U) when 1 mol 100^@C (Given molar enthalpy of vaporization of water at 1 bar and 373 k = 41 k J mol^(-1) and R= 8 cdot 3 J mol^(-1) K^(-1) ) will be

    A
    `3. cdot 7904 k J mol^(-1)`
    B
    `37. cdot 904 k J mol^(-1)`
    C
    `41. cdot 00 k J mol^(-1)`
    D
    `4. cdot 100 k J mol^(-1)`

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    A cylinder of fixed capacity 44.8 litres contains helium gas at standard temperature and pressure. What is the amount of heat needed to raise the temperature of the gas in the cylinder by 15.0^(@)C ? ( R= 8.31Jmol^(-1)K^(-1) )

    The ratio of specific heats of a gas in 1.40 . There are 200 moles of gas initially present. The gas undergoes adiabatic expansion and as a result which, the temperature of the gas falls from 400 K to 100 K. Give R=8.312J "mole"^(-1) K^(-1) . Calculate the amount or work done by the gas on the surroundings.

    The rate constant of a first order reaction at 300 K and 310 K are respectively 1.2 xx 10^(3) s^(-1) and 2.4 xx 10^(3) s^(-1) . Calculate the energy of activation. (R = 8.314 J K^(-1) mol^(-1))

    For the equilibirum, PCl_(5)(g) hArr PCl_(3)(g) + Cl_(2)(g) at 298 K, K = 1.8 xx 10^(-7) . Calculate Delta G^(@) for the reaction (R = 8.314 JK^(-1) mol^(-1)) .

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