Question

Multiple choice question

X, Y and Z are all gases that behave ideally and react according to the equation : X(g) + 2Y(g) = 2Z(g)

When 3.0 mol of X and 3.0 mol of Y are placed inside a container with a volume of 1.0 dm2 they react to form the maximum amount of Z.

the final temperature of the reaction vessel is 120 C
What is the final pressure inside the reaction vessel?

a) 4.49×10^8 Pa

b) 9.80×10^8 Pa

c) 1.47×10^7 Pa

d ) 1.96×10^7 Pa

Thanks.

Answer 1

Answer: The correct answer is Option c.

Explanation:

We are given:

Moles of X = 3.0 moles

Moles of Y = 3.0 moles

For the given chemical equation:

$X(g)+2Y(g)\rightarrow 2Z(g)$

By Stoichiometry of the reaction:

2 moles of Y reacts with 1 mole of X

So, 3.0 moles of Y will react with = $\frac{1}{2}\times 3.0=1.5mol$ of X

As, given amount of 'X' is more than the required amount. So, it is considered as an excess reagent.

Thus, 'Y' is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of Y produces 2 moles of Z

So, 3.0 moles of Y will produce = $\frac{2}{2}\times 3.0=3.0mol$ of X

Moles of X left in the vessel = 3.0 - 1.5 = 1.5 moles

To calculate the pressure, we use the equation given by ideal gas equation:

$PV=nRT$

where,

P = pressure of the gas

V = Volume of gas = $1.0dm^3$

n = number of moles = Moles of Z + Moles of X left = [3.0 + 1.5] = 4.5 moles

R = Gas constant = $8.31dm^3\text{ kPa }mol^{-1}K^{-1}$

T = temperature of the gas = $102^oC=[102+273]=393K$

Putting values in above equation, we get:

$P\times 1.0dm^3=4.5mol\times 8.31dm^3\text{ kPa }mol^{-1}K^{-1}\times 393K\\\\P=1.47\times 10^4kPa$

Converting the pressure in pascals, we use the conversion factor:

$1kPa=1000Pa=10^3Pa$

Converting into pascals, we get:

$\Rightarrow 1.47\times 10^4kPa\times \frac{10^3Pa}{1kPa}\\\\\Rightarrow 1.47\times 10^7Pa$

Hence, the correct answer is Option c.