As examined by Shannon in his paper "Entropy of printed English", higher order models which allow for dependencies between nearby characters are more accurate models of English. And in general, dependencies reduce but never increase entropy, so @Guut Boy is correct, the first shall have higher entropy. In the extreme case there is essentially zero entropy about the $(n+1)$st character if the $n$th character is q, it is almost always followed by u ("coq au vin" is taken to be French).
Edit To verify this experimentally given the string $(x_1,\ldots,x_n)$ over the alphabet $A$ define the sequence $(y_1,\ldots,y_{n-k+1})$ by
$$y_i=x_i,x_{i +1},\ldots,x_{i+k-1},\quad i =1,\ldots,n-k+1,$$ over $A^k.$
Compute the entropy corresponding to the frequencies of the $y_i$ sequence. Divide by $k$ to obtain entropy per symbol.
Put simply, $k=2$ would give the entropy of bigrams and divided by 2 it should be lower than that given by monogram frequencies.