a 2.50-mole quatity of NOCL was initially in a 1.50-L reaction chamber at 400 degrees celcius. after equilibrium was established, it was found that 28.0 percent of the NOCL had dissociated.
2NOCL(g) <---> 2NO(g)+ Cl2(g)
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What is the question?
2.5mole/1.5l=1.67M 28/100*167=0.46 0.46 the value of x and 27
mole of NOCL I divide x=0.46/2=0.23 Finaly Concentration (NOCL) =1.67-2(0.23)=1.21 and concentration of (NO) =2(0.23)=0.46 and concentration of Cl2=0.23 or kc=(0.46)^2(0.23)/(1.2)^2=3.402*10^-2
mole of NOCL I divide x=0.46/2=0.23 Finaly Concentration (NOCL) =1.67-2(0.23)=1.21 and concentration of (NO) =2(0.23)=0.46 and concentration of Cl2=0.23 or kc=(0.46)^2(0.23)/(1.2)^2=3.402*10^-2
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