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

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

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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) .

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.

If equilibrium constant K is 10^(3) , the Delta G^(theta) for the reaction at 300 K is (assume R = 8.314 JK^(-1) mol^(-1) ) :

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)) .

SUBHASH PUBLICATION-THERMAL PROPERTIES OF MATTER-FOUR MARKS QUESTIONS AND ANSWERS
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