Question

2 mole of zinc is dissolved in HCL at 25 degree Celsius calculate the work done in open vessel

Answer 1

Chemical equation :

ZN + 2HCl --->ZnCL2  + H2

If reaction is arried out in open reaction, then

External pressure = P=1 atmp

temp=T=25°c=25+273=298K

If V=0 [ because no gas is present initially]

Final volume occupied by 2 moles of Zn at 1 atmp is

PV=nRT

where R= universal constant=8.314 joule/kg

v=nRT/p

v=2x8.314x298/1

=4955.1445J

=4.9551455Kilo J

Workdone =W= -Px V

=- 1 x4.9551455KJ

=-4.9551455kJ

∴ the work done in open vessel is -4.9551455kJ

Answer 2

Given 2 moles of zinc is dissolved in hydrochloric acid at 25 degree celsius.

We can write the equation as

ZN + 2HCl --->ZnCL2  + H2

we get zinc chloride and hydrogen.

n = 2 moles

pressure = 1 atmosphere

T = 25 degree Celsius. This can be converted to kelvin by taking 273 + 25 = 298 kelvin .

R is called as universal constant and is given by the value 8.314 J

It is given that

Pv = nRT

v = nRT /P

v = 2 x 8.314 x 298 / 1

v = 4.955 kJ

Work done is given by - P x v

    = -1 x 4.955 = - 4.955 KJ