If 465 cm cube of sulphur(iv)oxide,can diffuse through porous partition in 30 seconds,how long will an equal volume , 620 cm cube of hydrogen sulphide take to diffuse through the same partition?(H=1, S=33, O=16)
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If 465 cm^3 of sulphur(IV) oxide, can diffuse through porous partition in 30 seconds, how long
will (a) an equal volume, (b) 620 cm^3 of hydrogen sulphide take to diffuse through the same
partition
Solution
To answer this question we should use Graham’s Law: rate of effusion for a gas is inversely
proportional to the square root of it’s molar mass:
Rate of effusion = 1/M.
For two gases:
Rate A/Rate B = MB/MA.
a) We have: V(SO2)= 0.465 dm3
t (SO2) = 30 s
V(H2S)= 0.465 dm3
Rate SO2 = n (SO2)/t;
n(SO2)= V/Vm; Vm =22.4 dm3
/mol;
n(SO2) = 0.465 dm3
/22.4 dm3
/mol =0.021 mol;
rate SO2= 0.021 mol/30 s =6.9210-4 mol/s.
Find rate H2S from Graham’s Law:
6.9210-4
/rate H2S = 34.09/64.07,
where M(H2S)=1.012+32.07=34.09 (g/mol), M(SO2) = 32.07+16.002 =64.07 (g/mol);
rate H2S = 9.4910-4 mol/s.
Find chemical amount of H2S:
n(H2S)= V/Vm; Vm =22.4 dm3
/mol;
n(H2S) = 0.465 dm3
/22.4 dm3
/mol =0.021 mol.
Rate H2S = n(H2S)/t;
9.4910-4 mol/s =0.021 mol/t;
t= 21.87 s.
b) We have: V(SO2)= 0.465 dm3
t(SO2)= 30s
V(H2S) = 0.620 dm3
Rate SO2 = n (SO2)/t;
n(SO2)= V/Vm; Vm =22.4 dm3
/mol;
n(SO2) = 0.465 dm3
/22.4 dm3
/mol =0.021 mol;
rate SO2= 0.021mol/30 s =6.9210-4 mol/s.
Find rate H2S from Graham’s Law:
6.9210-4
/rate H2S = 34.09/64.07,
where M(H2S)=1.012+32.07=34.09 (g/mol), M(SO2) = 32.07+16.002 =64.07 (g/mol);rate H2S = 9.4910-4 mol/s.
Find chemical amount of H2S:
n(H2S)= V/Vm; Vm =22.4 dm3
/mol;
n(H2S) = 0.620 dm3
/22.4 dm3
/mol =0.0277 mol.
Rate H2S = n(H2S)/t;
9.4910-4 mol/s =0.0277 mol/t;
t= 29.19 s.
Answer: a) 21.87 s; b) 29.19 s