W(g) + x(g) --> y(g) +z(g)
gases w and x react in a closed, rigid vessel to form gases yand z according to the equation above. the initial pressure of w(g)is 1.20 atm and that of x(g) is 1.60 atm. no y(g) or z(g) isinitially present. the experiment is carried out at constanttemperature. what is the partial pressure of z(g) when the partialpressure of w(g) has decreased to 1.0 atm?
a) 0.20 atm
b) 0.40 atm
c) 1.0 atm
d) 1.2 atm
e) 1.4 atm

A) 0.20 atm

Explanation:

Let's consider the following reaction.

W(g) + X(g) ⇄ Y(g) +Z(g)

We can find the partial pressure of Z at a time that we will call the equilibrium using an ICE chart. We identify 3 stages: Initial, Change and Equilibrium and complete each row with the pressure or change in pressure.

        W(g) + X(g) ⇄ Y(g) +Z(g)

I        1.20    1.60       0      0

C        -x        -x         +x     +x

E     1.20 -x  1.60 -x    x        x

At equilibrium, the partial pressure of W is 1.0 atm.

1.20 atm - x = 1.0 atm

x = 0.20 atm

The pressure of Z is x = 0.20 atm

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.

i wish i could but this one is weird give me abc and i can solve it.