Could someone explain? I’ve tried for almost 2hours and can’t figure out what to do. Thanks
It's moles. The reaction can be expressed:
acid + alcohol <==> ester + water
The initial moles are in the question. Once equilibrium is established 'x' mol of the reactants change and produce 'x' mol of the products. So the equilibrium moles = acid = (0.105 - x) alcohol = (0.080 - x) ester = x water = (0.111 + x)
The total mol acid at equilibrium = acid catalyst + acid remaining = 0.00400 + (0.105 - x)
The initial moles are in the question. Once equilibrium is established 'x' mol of the reactants change and produce 'x' mol of the products. So the equilibrium moles = acid = (0.105 - x) alcohol = (0.080 - x) ester = x water = (0.111 + x)
The total mol acid at equilibrium = acid catalyst + acid remaining = 0.00400 + (0.105 - x)
can you take it from here?
I’m pretty sure 2b(i) is about the reaction of hydrogen and iodine. I don’t think that’s the question the OP wanted help with, but it’s a useful start nonetheless.
I’m pretty sure 2b(i) is about the reaction of hydrogen and iodine. I don’t think that’s the question the OP wanted help with, but it’s a useful start nonetheless.
For each value of a in the table, try adding [I2] and y and see if you notice anything.
I’m pretty sure 2b(i) is about the reaction of hydrogen and iodine. I don’t think that’s the question the OP wanted help with, but it’s a useful start nonetheless.
Could someone explain to me why you cannot simply rearrange y= a√kc / 2+ √kc to y / a = √kc / 2+ √kc, and as √kc / 2+ √kc is a constant, you can use the values for a and y to find out the value of this constant and use this to find the other values? if that makes any sense?
Could someone explain to me why you cannot simply rearrange y= a√kc / 2+ √kc to y / a = √kc / 2+ √kc, and as √kc / 2+ √kc is a constant, you can use the values for a and y to find out the value of this constant and use this to find the other values? if that makes any sense?
I think you could but that’s a really long method for just 2 marks.