For a chemical reaction
4A + 3B 6C + 9D
Rate of formation of C is 6 102 mol L1 s1 and rate of disappearance of A is 4 102 mol L1 s1. The rate of reaction and amount of B consumed in interval of 10 seconds, respectively will be :
- A10 102 mol L1 s1 and 30 102 mol L1
- B1 102 mol L1 s1 and 30 102 mol L1
- C10 102 mol L1 s1 and 10 102 mol L1
- D1 102 mol L1 s1 and 10 102 mol L1
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Correct answer: B
4A + 3B $\to$ 6C + 9D
Rate of reaction = $${{ - d[A]} \over {dt}} \times {1 \over 4} = {{ - d[B]} \over {dt}} \times {1 \over 3} = {{ + d[C]} \over {dt}} \times {1 \over 6} = {{ + d[D]} \over {dt}} \times {1 \over 9}$$
Rate of reaction $$ = {{ + d[C]} \over {dt}} \times {1 \over 6} = {{6 \times {{10}^{ - 2}}} \over 6} = {10^{ - 2}}$$ mol L$-$1 s$-$1
Rate of reaction $$ = {{ - 1} \over 3}{{d[B]} \over {dt}}$$
$${{ - d[B]} \over {dt}} = 3 \times $$ rate of reaction $$ = 3 \times {10^{ - 2}}$$ mol L$-$1 s$-$1
After interval of 10 sec. $$ = 3 \times {10^{ - 2}} \times 10$$
$$ = 30 \times {10^{ - 2}}$$ mol L$-$1
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