photochemical decomposition of HI show that -d[HI]/dt=2Ia
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Reaction----->
2HI-----> H2 + I2
-1/2d(HI)/dt = d(H2)/dt = d(I2)/dt
I.e rate of decomposition of HI= rate of formation of H2=rate of formation of I2
2HI-----> H2 + I2
-1/2d(HI)/dt = d(H2)/dt = d(I2)/dt
I.e rate of decomposition of HI= rate of formation of H2=rate of formation of I2
sejaljladhava:
Plz give me long answer because my assignment question
time limit is gone
u can include this as well before my answer started
m1A+m2B----)nC
for one mole-----). Rate =-1/m1 d(A)/dt=-1/m2 d(B) /dt=1/n d(C)/dt
Ok thankx
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