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The following reaction systems are at equilibrium in separate sealed containers - VCE - SSCE Chemistry - Question 7 - 2010 - Paper 1
Question 7
The following reaction systems are at equilibrium in separate sealed containers. The volumes of the containers are halved at constant temperature. Which reaction ha... show full transcript
Worked Solution & Example Answer:The following reaction systems are at equilibrium in separate sealed containers - VCE - SSCE Chemistry - Question 7 - 2010 - Paper 1
Step 1
Identify the Reaction for A. N2(g) ↔ 2NO(g)
Answer
For this reaction, halving the volume results in an increase in concentration of both N2 and NO. The equilibrium shifts to the right, producing more NO. The percentage change in concentration can be calculated as follows:
Initial concentration changes from x to 2x after halving the volume, resulting in:
Percentage change = ( \frac{(2x - x)}{x} \times 100 = 100% )
Step 2
Identify the Reaction for B. H2(g) + I2(g) ↔ 2HI(g)
Answer
For this reaction, both reactants and products will experience a change in concentration. The equilibrium will shift to the right due to a decrease in volume, leading to:
Percentage change for H2 and I2 would be similar as they are both 1:1 ratios being converted to 2 units of HI. Thus:
Percentage change = ( \frac{2y - y}{y} \times 100 = 100% )
Step 3
Identify the Reaction for C. 2CO(g) ↔ 2C(0)(g) + O2(g)
Answer
In this scenario, the reaction has 2 moles of gas on each side (2CO ↔ 2C + 1O2). When the volume is halved, the pressure increases and the system will not favor either side significantly, resulting in a negligible percentage change.
Thus, we would calculate a small percentage change.
Step 4
Identify the Reaction for D. CO(g) + 2H2(g) ↔ CH3OH(g)
Answer
In this case, there are 3 moles of reactants (1 CO + 2 H2) and only 1 mole of product (1 CH3OH). Halving the volume greatly increases the reactant concentrations, making the equilibrium shift left.
The percentage change can be calculated as:
- Initial for CO would be x, and after halving, it goes to 2x, while H2 goes from y to 2y. Thus: Percentage change = ( \frac{(2x - x) + (2y - y)}{x+y} \times 100)