1.

At 380^(@)C, the half-life period for the first order decomposition of H_2O_2 is 360 minute. The energy of activation of the reaction is 200 kJ mol^(-1). What will be the time required for 75% decomposition at 450^(@) C?

Answer»

20.39 min
30.03 min
1223.4 s
2000 s

Solution : `K_1=(0.693)/(t_(1//2))=(0.693)/(360) = 1.925xx10^(-3) min^(-1)`
`log_(10)(k_2)/(k_1)=(E_a)/(2.303R)((T_2-T_1)/(T_1T_2))`
`log_10(k_2)/(1.925xx10^(-3))=(200xx10^3)/(2.303xx8.314)[(723-653)/(653xx723)]`
`K_2`=0.068 `min^(-1)`
`t=(2.303)/(k_2) log_(10)(a)/(a-x) implies (2.303)/(0.068) log_(10) (100)/(25)`
`THEREFORE` t=20.39 min =1223.4 s


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