Question

2PbS(s) + 3O2(g)  2PbO(s) + 2SO2(g) H = -827.4 kJ What is the volume of sulfur dioxide produced at STP if 975 kJ of heat are liberated?

Answer 1

Answer:

$V=52.8L$

Explanation:

Hello,

In this case, given the reaction:

$2PbS(s) + 3O_2(g) \rightarrow 2PbO(s) + 2SO_2(g)$

827.4 kJ of heat are releasing 2 moles of sulfur dioxide, thus, for 975 kJ we find the following released moles:

$n=\frac{-975 kJ*2mol}{-827.4kJ} =2.36mol$

Then, at STP conditions (273K and 1atm) we compure the volume by using the ideal gas equation:

$V=\frac{nRT}{P}=\frac{2.36mol*0.082\frac{atm*L}{mol*K}*273K}{1atm} \\\\V=52.8L$

Regards.

Answer 2

The volume of sulphur dioxide, SO₂ produced at STP if 975 KJ of heat are liberated is 52.9 L

We'll begin by calculating the number of mole of SO₂ produced when 975 KJ of heat are liberated. This can be obtained as follow:

2PbS + 3O₂ —> 2PbO + 2SO₂ ∆H = –827.4 KJ

From the balanced equation above,

827.4 KJ of heat energy produced 2 moles of SO₂.

Therefore,

975 KJ of heat will produce = (975 × 2)/ 827.4 = 2.36 moles of SO₂.

• Finally, we shall determine the volume of SO₂ produced from the reaction. This can be obtained as follow:

Number of mole (n) = 2.36 moles

Pressure (P) = STP = 1 atm

Temperature (T) = STP = 273 K

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) =?

PV = nRT

1 × V = 2.36 × 0.0821 × 273

V = 52.9 L

Thus, the volume of sulphur dioxide, SO₂ produced at STP if 975 KJ of heat are liberated is 52.9 L

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